Rotational Dynamics - NetBadi.com

ROTATIONAL DYNAMICS

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Illustration: 1

A one – miece cylinder is shaped as in figure with a core section producing from the larger drum. The cylinder is free to rotate around the central axis shown in the drawing. A rope wrapped around the drum, of radius R₁, enters a force T₁ to the right on the cylinder. A rope warped around the core, of radius R₂, enters a force T₂ downward on the cylinder. (a) what is the net torque acting on the cylinder about the rotation axis (which is the z – axis in figure)? (b) suppose T₁ = 5N, R₁ = 1.0 m, T₂ = 6n, and R₂ = 0.50 m. what is the net torque about the rotation axis and which way does the cylinder rotate if it starts from rest

Figures

Solution:

(a) The torque due to T_i is – R₁ T₁. It is negative because it tends to produce a clockwise rotation from the point of view in figure. The torque due to T₂ is T₂ = R₂ T₂ and is positive because it tends to produce a counter clockwise ? Wnet W1W2 R T2 2R T1 1

(b) Wnet R T2 2R T1 1

= (0.5) (6) – (1) (5) = - 2 N.m

Because the net torque is negative, the cylinder rotates clockwise from rest.

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Illustration: 2

A triangular plate of uniform thickness and density is made to rotate about an axis perpendicular to the plane of the paper and (a) Passing through A (b) passing through D by the application of the same force F at c (mid – point of AB) as shown in fig. In which case the torque is more?

Solution: 2 2 A D l l F and F W § ·_{¨ ¸} W  § ·_{¨ ¸} © ¹ © ¹

? Both are equal in magnitude Illustration: 3

A rectangular plate in the x – y plane is shown in the figure. A force F = 30 N is applied at point B. Find the moment of F about (a) the origin O (b) the point C (c) x – axis, y – axis, and z – axis. T300

Solution :

(a) W0 Fcos 30 2u Fsin 30 1u

3 1 30 2 30 2 2 § ·  u  _{¨ ¸} © ¹ 1 30 3 . 2 N m § ·  _¨  _¸ © ¹ (b) Wc Fcos 2u 30 3Nm 0 0 x and y W W 1 30 3 . 2 z N m W  §_¨  ·_¸ © ¹ Illustration: 4

A force F 2i3j4k N is applied to a point having position vector r 3i2jk m. Find the torque due to the force about the axis passing through origin. Figure

Solution: r F W u 3i2jk u 2i9j4k

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3 2 1 2 3 4 i j l 5i 10j 5k Nnm W ?   2 2 2 5 10 5 5 6Nm. W ?    SOLVED EXAMPLES:

1. A 450 N vertical force is applied to the end of a leaves which is attached to a shaff at 0. Determine (a) the torque of the 450 N force about 0; (b( the horizontal force applied at A which create the same torque ‘0’. (c) the smallest force applied at A which creates the same torque about 0; (d) how four from the shaff a 1100 N vertical force must act to create the same torque about 0; (e) whether any one of the force obtained in parts b, c and d is equivalent to the original force.

Figures Solutions:

a. Torque about 0. The A distance from 0 to the line of action of 450 N force is d = 0.6 cos 60 = 0.3 m. The torque about 0 of the 450 N force is W0 F d 450 0.3 {135Nm. Since the force tends to rotate the leves clockwise

about 0, the torque will be represented by a vector _W0 perpendicular to the plane of the figure and pointing into the paper

b. Horizontal force d = 0.6 sin 60 = 0.5 m; W0 Fd Ÿ 135 F 0.52 Ÿ F 260N.

c. Similar force, since W0 Fd, the smallest value of F occurs when d is maximum. We chose the force perpendicular

to OA and rote that d = 0.6 m. W0 Fd - 135 = - F (0.6) Ÿ F = 225 N.

d. 1100 N vertical force 135 1100d d 0.12m.   Ÿ cos 60 0.24 . OB d OB m ? Ÿ

e. None of the forces considered in parts b, c and d is equivalent to the original 450 N force. Although they hence the same torque about 0, they have different x and y components. In of her words, although each force mannes, each causes the leves to pull on the shaft in a different way.

2. A particle of mass m is projected with a speed u at an angle T with the horizontal at time t = 0. Find the torque of the weight of the particle about the point of projection 0 at time t if when the particle is at the highest point. Solution: W0 W

= - W cos T t at the highest point,

0 -sin i v u T sin u t g T Ÿ 0 sin cos .u W u g T W T ? 

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2

sin cos

m u T T



Since the force tends to rotate the leves clockwise about 0, the torque will be represented by a vector W0

perpendicular to the plane of the projectile and pointing into the papers.

3. A simple pendulum of length l is pulled a side to make an angle T with the vertical. Find the magnitude of the torque of weight W of the bob about the point of suspension. When is the torque zero.

Solution: Figure (i) W0 wlsinT sin wl T (clockwise) (ii) W0 0 Ÿ w lsinT 0 0. . .,i e T Ÿ

? When the bob is as the lowest point.

4. Calculate the total torque acting on the body shown in figure, about the point 0. Solution: W0 10 4 5 0 Figure

+ 20 cos 6 (4) + 15 sin 37 (6) = - 40 + 40 + 54 = - 54 N cm = 0.54 N.

5. A cord is wrapped around a homogenous dist of radius r = 0.5 m and mass m = 15 kg. If the cord is pulled upward with a force T = 180 N, determine the torque at the center

Solution: W0 Tr 180 0.5u = - 90 N. m.

= 90 N. m. (clockwise) figure

TORQUE AND THE VECTOR PRODUCT:

Recall our stationary exercise bicycle from the preceding section. We caused the rotational motion of the wheel by applying forces to the pedals. When a force is exerted not pass through the pivot, the body tends to rotate about that axis. For example, when you push on a door, the door rotates about an axis through the hinges. The tendency of a force to rotate a body about some axis is measured by a vector quantity called torque. Torque is the cause of changes in rotational motion, and is analogous to force, which causes changes in translational motion. Consider the wrench pivoted about the axis through O in figure 10.11 The applied force F generally can act at an angle I with respect to the position vector r locating the point of application of the force. We define the torque W (Greek letter tau) resulting from the force F with the expression † W{r FsinI.

It is very important to recognize that torque is defined only when a reference is specified, from which the distance r is determined. We can interpret. Equation 10.18 in two different ways. Looking at the force components in figure

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10.11, we see that the component F cos I parallel to r will not cause a rotation around the pivot point, because its line of action passes right through the pivot.

The line of action of a force is imaginary line colinear with the force vector and extending to infinity in both directions. In general, torque is a vector. For rotation about a fixed axis, however, we will use italic, non bold notation and specify the direction with a positive or a negative sing, as we did for angular speed and acceleration in Section 10.1. We will treat the vector nature of briefly in a short while.

THE RIGID BODY IN EQUILIBRIUM:

We have defined a rigid body and have discussed torque as the cause of changes in rotational motion of a rigid body: We can now establish models for a rigid body subject to torques that are analogous to those for a particle to forces. We begin by imagining a rigid body with balanced torques.

Consider two forces of equal magnitude and opposite direction applied to an object s shown in figure 10.16 a. The force directed to the right tends to rotate the object clockwise about an axis perpendicular to the diagram through O, whereas the force directed to the left tends t rotate it counterclockwise about that axis. Because the forces are of equal magnitude and act at the same perpendicular distance from ), their torques are equal in magnitude. Thus, the net torque on the rigid body is zero. The situation shown in figure 10.16 b is another case in which the net torque about O is zero (although the net force on the object is not zero), and we can devise many more cases.

With no net torque, no change in rotational motion and the rotational motion of the rigid body remains in its original state. This is an equilibrium situation; analogous to translational equilibrium, discussed in Chapter 4. We now have two conditions for complete equilibrium of an object, which can be stated as follows:

The net external force must equal zero

0

F

¦

The net external torque must be zero about any axis

0 W

¦

The first condition is a statement of translational equilibrium. The second condition is a statement of rotational equilibrium. In the special of static equilibrium, the object is at rest, so that it has no translational or angular speed

. ., CM 0 0

i e X and Z .

The two vector expressions given by Equations 10.24 and 10.25 are equivalent, in general, to six scalar equations: three from the first condition of equilibrium, and three from the second (corresponding to x, y, and z components). Hence, in a faced with solving a set of equations with many unknowns. Here, we restrict our discussion to situations in which all the forces lie in the x, y plane. (Forces whose vector representations are in the same plane are said to be coplanar.) With this restriction, we need to deal with only three scalar equations. Two of these come from balancing the forces in the x and y directions. The third comes from the torque equation, namely, that the net torque about an axis through any point in x y plane must be zero. Hence, the two conditions of equilibrium provide the equations.

0 0 0

x y z

F F W

¦

¦

¦

Where the axis of the torque equation is arbitrary.

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Quick Quiz 10.8:

a. Is it possible for a situation to exist in which Equations 10.24 is satisfied, but Equation 10.25 is not? b. Can Equation 10.25 be satisfied, but Equation 10.24 is

In working static equilibrium problems, it is important to recognize all external forces acting on the object. Failure to do so will result in an incorrect analysis. The following procedure is recommended when analyzing an object in equilibrium under the action of several external forces:

Problem – Solving Strategy:

1. Make a sketch of the rigid body under consideration.

2. Draw a free – body diagram, and label all external forces acting on the object. Try to guess the correct direction for each force. If you select a direction that leads to a negative sign in your solution for a force, do no be alarmed; this merely means that the direction of the force is the opposite of what you guessed.

3. Resolve all forces into rectangular components, choosing a convenient coordinate system. Then apply the first condition for equilibrium, Equation 10.24. Remember to keep track of the signs of the various force components. 4. Choose a convenient axis for calculating the net torque on the rigid body. Remember that the choice of axis for the torque equation is arbitrary therefore; choose an axis that will simplify your calculation as much possible. Usually, the most convenient axis for calculating torques is one through a point at which several forces act, so that their torques around this axis is zero. If you don’t know a force or don’t need to know a force acts. Apply the second condition for equilibrium, Equation 10.25

5. Solve the simultaneous equations for the unknowns in terms of the known quantities.

Example 10.8 Standing on a Horizontal Beam :

A uniform horizontal beam of length 8.00 m and weight 200 N is attached to a wall by a pin connection. Its far end is supported by a cable that makes an angle of 53.00 _{with the horizontal (Fig. 10.17 a). If a 600 – N man stands} 2.00 m from the wall, find the tension in the cable and the force exerted by the wall on the beam.

Solution: The beam – man system is at rest and remains at rest, so it is clearly in static equilibrium. First we must identify all the external forces acting on the system, which we do in the free – body diagram in figure 10.17 b. These are the weights of the beam and the man, the force T exerted by the cable, the force R exerted by the wall at the pivot (the direction of this force is unknown). (The force between the man and the beam is internal to the system, so it is not included in the free – body diagram.) Notice that we have imagined the gravitational force on the beam as acting at its center of gravity. Because the beam is uniform, the center of gravity is at the geometric center. If we resolved T and R into horizontal and vertical components (Figure 10.17 c) and apply the first condition for equilibrium for the system, we have

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1.

¦

2.

¦

Because we have unknowns – R, T and T - we can obtain a solution from these two expression alone.

To generate a third expression, let us invoke the condition for rotational equilibrium, because the system can be modeled as a rigid body in equilibrium. A convenient axis to choose for our torque equation is the one that passes through the pivot at the wall. The feature that makes this point so convenient is that the force R and the horizontal component of T both have a lever arm of zero, and hence zero torque, about this pivot. Recalling our convention for the sign of the torque about an axis and noting that the lever arms of the 600 – N, 200 N, and T sin 530 _{forces are 2.00 m, 4.00 m, and 8.00 m, respectively, we have}

0 sin 53.0 8.00 600 2.00 T m N m W 

¦

The torque equation gives us one of the unknowns directly! This is due to our judicious choice of the axis. This value is substituted into (1) and (2) to give

cos 188

R T N

sin 550

R T N

We divide these two equations to find

550 tan 2.93 188 N N T 0 7.1 T Finally, 0 188 188 581 cos cos 71.1 N N R N T .

If we had selected other axis for the torque equation, the results would have been the same, although the details of the solution would be some what different. For example, if we had chosen to have the axis pass through the center of gravity of the beam, the torque equation would involve both T and R. However, this equation, coupled with (1) and (2), could still be solved for the unknowns T, R and T, yielding the same results. Try it!

Example 10.9 The Leaning Ladder:

A uniform ladder of length l and mass m against a smooth, vertical wall (Fig. 10.18 a). If the coefficient of static friction between ladder and ground is P_s 0.40, find the minimum angle T_min such that the ladder does not slip.

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Solution: The ladder is at rest and remains at rest, so we model it as a rigid body in equilibrium. The free – body diagram showing all the external forces acting on the ladder is illustrated in Figure 10.18 b. The reaction R exerted by the ground on the ladder is the vector sum of a normal force n and the force of static friction f. The reaction force P exerted by the wall on the ladder is horizontal, because the wall is smooth, meaning that it is frictionless: Thus, P is simply the normal force on the ladder from the wall. From the first condition of equilibrium applied to the ladder, we have

0 x F  f P

¦

¦

We see from the second equation that n = mg. Further moment when the ladder is on the verge of slipping, the force of friction must be a maximum, given fs, max Psn. The find T, we use the second condition of equilibrium. When the torques are taken about the origin O at the bottom of the ladder, we have

sin cos 0 2 O P mg W T T

¦

This expression gives

min .max 1 tan 1.2 2 2 s 2 s 2 0.40 mg n n P f n T P 0 min tanT 51 .

It is interesting that the result does not depend on l or m. The answer depends only on Ps.

EXERCISE: Two children weighing 500 N and 350 N are on uniform board weighing 40.0 N supported at its center (“see – saw”). If the 500 N child is 1.50 m from the center, termine (a) the upward force exerted on the board by the support and (b) where the 350 – N child must be balanced the system.

Answer: (a) 890 N (b) 2.14 m from the center.

Notes:

1. On tilting, a body will restore its initial position due to torque of weight about the point O hill the line of action of weight passes through its base. On tilting, a body will topple due to torque of weight about 0, if the line of action of weight does not pass through its base. This is self evident from Figures

2. A body is said h be in rotational equilibrium if resultant torque acting on it is zero, i.e., 6 W 0 e.g., in case of

beam balance or see – saw system will be in rotational equilibrium if

1 2 0

W W

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Or F l_{1 1}0 F l _{2 2} 0

Or F l1 1 F l2 2

However, if W1!W2 L.H.S will none downwards and if W1W2 R.H.S will none downwards and

the system will not be in rotational equilibrium. Figure.

Illustration – 1

Discuss whether or not a body is in equilibrium and what type of motion does it execute when two oppositely directed forces are applied to it.

Solution: There are four possibilities depending on the magnitude and point of applications of the forces:

(A) When the forces are equal and act along the same line. As shown in figure (a) in this situation:

0 0

F and W

6 6

So the body is in both translator as well as in rotatory equilibrium, i.e., body will remain as rest initially it was at rest. (B) When the forces are equal and does not act along the same line.3

As shown in figure (b) in this situation 6 F 0 but 6 z W 0 Fl so the body is the translatory equilibrium but not in rotatory equilibrium, i.e., if will not translate but will rotate, i.e., will spin.

(C) When the forces are unequal and act along the same line. As shown in figure (c) in this situation.

0but W 0

6 z 6

So the body i.e, in rotatory equilibrium but not in translatory equilibrium, i.e., it will not rotate but will translate, i.e., will slip or said

(D) When the forces are unequal and does not act along the same line. As shown in figure (D) in this situation:

0 0

F and W

6 z 6 z

So the body is neither in translatory nor in rotatory equilibrium, i.e., it will rotate and as well as translate, i.e., will roll.

Examples:

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1. A uniform cube of side 2a and mass M rests on a rough horizontal table. A horizontal force F is applied normal to one of the faces at a point that is directly above the center of the face, at a height (c) above the base. What is the minimum value of F for which the cube begins to tip about an edge?

Solution: As shown in Fig, the cube will tip about the edge through ‘O’ if torque of F exceeds the torque of Mg (torque of R about O is zero) i.e., F Cu !Mg a or F Mga

c

! _i.e., F_min aMg

c .

Case (i) Let 3

2 a

C .

2. In many situations an external force is applied to a body cause if to slide along a surface. In certain cases, the body may tip over before sliding ensues; show the criteria that may be used to predict whether sliding or tipping occurs.

Solution: Fig (a) shows a body which rests on a horizontal plane. The weight of the body is W and it has the dimensions 2b and h, P is the coefficient of friction. A force U, at a distance c above the plane, is applied to the body.. Three figures

The free body diagram of the body is shown in figure (b). The normal reaction force N is assumed to act at distance of d to the right of the line of action of the weight W. For this first case, the body is assumed not to slide.

0; 0 A PC Wd W 6   Wd C p Ÿ

As the force P is applied at larger values of h, the location of the line of action of the normal force N moves to the right. A limiting condition occurs when d = b. For this case the normal force acts at the right – hand edge of the body. The body is in a condition of impending tipping motion about the right hand corner, and the corresponding value of C is desegregated as C_man.

0 0; PCman Wb 0 W 6   man Wd C p ……… (1).

The condition for sliding:

It sliding motion of the body is assumed to be impending, the friction force F_man is given by Fman PX . The equilibrium requirements of the body are

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0; 0 x F U PX 6 0, 0 y F W X 6  

The force P is then formed to be

W

U P ……… (2).

P in the above equation is the value of applied force which is required to cause impending sliding motion of the body. If sliding and timing are assumed to be equally likely to occur, the W may be eliminated between Equations (1) and (2), with the result

man

b C

P ……… (3).

The result in the above equation is independent of the weight W and height h of the body, and of the applied force P, and is a function only of the dimension b and the coefficient of friction.

From Equation (3) it is seen that if we have a block of such proportions that b/h > m, the block will slide before overturning, regardless of the point of application of the horizontal force P.

The five possible district regious of motion are:

1. C < C_man U P w; Neither tipping nor sliding occurs, and the body remains at rest. 2. C < C_man, U Pw. The body does not tip, and sliding motion is impending.

3. C < C_man U Pw; The body does not tip but sliding motion occurs.

4. C = C_man, U Pw; Both sliding and tipping motion are impending simultaneously..

5. C > C_man : The body will tip over for any value of P, with one or the other of the following conditions. If

w U Pt

or U Pd w, the on set of tipping motion will occur with the tipping edge remaining stationary with respect to the

surface. Otherwise with

w U

P or U P! w, the onset of tipping motion will be accompanied by the onset of sliding motion of the edge along the surface.

3. A cubical block of mass m and edge a slides down a rough inclined plane of inclination T with a uniform speed. Find the torque of the normal force acting on the block about its center.

Solution : Since the block slides with constant speed, so the net force on the block must be zero.

0; sin x F f mg D ? 6 Figure 0; cos y F X mg D 6

Now, 6Wc 0, (rotational equilibrium torque due to normal reaction (clockwise)

= torque due to frictional force (anti clockwise)

= sin 2 a

mg D .

4. Two small kids weighing 10 kg 15 kg are tying to balance a see saw of total length 5m, with the fulcrum of the centre. If one of the kid is sitting at an end, where should the other sit.

Solution: It is clear that the 10 kg kid should sit at the end and the 15 kg kid should sit closer to the centre. For rotational equilibrium, 6Wc 0 (10 kg)g (2.5 m) – (15 kg) g (x) = 0 1.7 . x m Ÿ

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5. The ladder shown in figure has negligible mass and rests on a frictionless floor. The cross bar connects the two legs of the ladder at the middle. The angle between the two legs is 600_{. The fat person sitting on the ladder has} a mass of 80 kg. Find the contact force exerted by the floor on each leg and the tension in the cross bar. (g = 10) m/s2_.

Solution: The forces acting on different parts are shown in figure. The body is in equilibrium.

0; 2 0 y F W X 6 †  80 10 400 2 2 W X u X Ÿ

Next consider the equilibrium of the left leg of the ladder l _{cos 30}2

0; sin 30 1 0 c l T W X 6  2 tan 30 400 2 tan 30 T X Ÿ u u 800 3 T X Ÿ

Rigid Body under a net torque:

In the preceding section, we investigated the equilibrium situation in which the net torque on a rigid body is zero. What if the net torque on a rigid body is not zero? In analogy with Newton’s second law for translational motion, we should expect the angular speed of the rigid body to change. The net torque will cause angular acceleration of the rigid body.

Let us imagine a rotating rigid body again as a collection of particles. The rigid body will be subject to a number of forces applied at various locations on the rigid body, at which individual particles will be located. Thus, we can imagine that the forces on the rigid body are exerted on individual particles of the rigid body object due to the torques resulting from of these forces around the rotation axis of the rotating body. Any applied force can be represented by its radial component and its tangential component. The radial component for an applied force provides t\no torque because its line of action goes tangential component of an applied force contributes to the torque.

On any given particle, described by index variable i , within the rigid body, we can use Newton’s second law to describe the tangential acceleration of the particle;

i

t i i

F m at

Where the t subscript refers to tangential components.

Multiply both sides of the expression by r_i , the distance of the particle from the rotation axis;

i i i t i i t r F r m a i i r Fi t W  2 i m ri i i W D ati riDi 

Add the torques n all particle of the rigid body

2 . . i i i i i i m r W D

¦ ¦

The left side is the net torque on all particles of the rigid body. The net torque associated with internal forces is zero, however. To understand this, recall that Newton’s third law tells us that the internal force occur in equal and opposite pairs that lie along the line of separation of each pair of the particles. The torque due to each action

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– reaction force pair is therefore zero. On summation of all torques, we see that the net internal torque vnishes. The term on the left, then reduce to net central torque. On the right, we impose the rigid body model by demanding that all particles have the same angular acceleration.

2 . i i m r W § · D Ÿ 6 ¨_¨ ¸_¸ ©

¦

i.e., The net torque acting on the rigid on the rigid body is proportional to its angular acceleration proportional constant is the moment of inertia.

Problem 27: A uniform cylinder of radius R is spinned about its axis with an angular velocityZ0 and then placed into

a corner as shown in figure 48. The coefficient of friction between the wall and cylinder as well as the ground and the cylinder is P. How many turns will the cylinder accomplish before it stops?

Solution: First Method: Free body diagram is shown in figure 49. Its is clear from figure that

2 1 1 N F PN ……… (1) And F₂ + N₁ = mg ……… (2) Now, 2 2 2 1 F PN P N ……… (3) From equation (2), 2 2 2 1 1 1 1 1 / 1 N mgP N or N P mg or N mg P 2 2 2 2 1 1 / 1 2 1 / 1 F PN Pmg P and F P N P mg P ?   …………(5)

Total frictional force acting on the cylinder : 1 2 ₂

1 1 m g F F F P P P   

Torque due to this force, 2

1 1 m g R F R P P W P   As 2 2 2 1 2 1 1 , 1 1 1 / 2 1 m g R g m R R P P P P W W D D P P   ?   .

30.13 Motion of a Body Tied to a String and Passing Over a Pulley

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Consider a body tied to string and passing over a pulley as shown in figure. 30.8. Let the body descend under the force of gravity.

Let m = Mass of the body in kg (so that its weight is m. g Newton’s) M = Mass of the pulley in kg

I = Moment of inertia of the pulley, r = Radius of the pulley,

k = Radius of gyration of the pulley. a = Linear acceleration of the body,

D = Angular acceleration of the pulley, and P = Pull in the string in Newton’s.

+

P

P m

Kinetic of Motion of Rotation:

First of all, consider the motion of the body, which is coming down. We known that the forces acting on it are m. g Newton’s (downwards) therefore resultant force acting on it

= m. g – P Newton’ s. …………. (i)

Since the body is moving downwards, with an acceleration (a) therefore force acting on it. = m. a Newton’s …………. (ii)

Equating equations (i) and (ii),

m.g – P = m. a …………. (iii)

now consider motion of the pulley, which is rotating about its axis due to downward motion of the body tied to the string. We know that linear acceleration of the body is equal to the angular acceleration of the pulley.

. r

D D

? …………. (iv)

And torque, T = Tension in the string

u Radius of the pulley

= P u r ……… (v)

We also know that torque on the pulley, T = I. D

Equating equations (v) and (vi)

. P ru ID

2

. .

P IDr …………. (Multiplying both sides by r}

= I. a 2 . I a P r ? _{……… (vii),}

Substituting the value of P in equation (iii).

2 . . I a . m g m a r  2 . . I a . m g m g r 

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2 . I a m m g r § _ · ¨ ¸ © ¹ 2 2 ₂ . . a m g a I _{M k} m _m r _r ? § _ · § ·  ¨ ¸ ¨_¨ ¸_¸ © ¹ _{© ¹} ……… (viii)

Now substituting the value of a in equation (vii),

2 2 2 . . . I m g I m g P I r _m m r I r u § _ ·  ¨ ¸ © ¹ ……….. (ix)

Problem 22: A light thread with a body of mass m tied to its end is wound on a uniform solid cylinder of mass M and radius R (fig 42). At a moment t = 0, the system is set in motion. Assuming the friction in the axle of the cylinder to be negligible, find the time dependence of (a) the angular velocity of the cylinder.

Solution: (a) Because the system is released from rest hence u = 0 at t = 0. According to free body diagram shown in figure 43, we have

mg = T = ma ………… (1) figure

Where a represents the downward acceleration of the system and T the tension in the thread. If I is the moment of inertia of the cylinder about an axis passing through its centre and D , the angular acceleration produced in it, then

2

1 1

2 2

T R ID M R D or T M RD ………… (2) figure

Adding (1) and (2), we get : 1 1 1 ( / )

2 2 2

mg ma M RDmDR M RD M RD asD a R

Or, 2mg 2m RM R D or, D _2m2_mg_{M R} .

Integrating on both sides, we get: Z _2m2m g t_{M R}C Now, at t = 0, Z = 0, Hence C = 0

Thus Z _2m2m g t_{M R}.

Problem 23: A wheel of radius r and moment of a inertia I about its axis is fixed at the top of an inclined plane of inclination T (Fig 44). A string is wrapped round the wheel and its free end supports a block of mass M which can slide on the plane. Initially, the wheel is rotating at a speed Z in a direction such that the block slides up the plane. How far will the block move before stopping? Figure

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Solution: As the block moves up, it goes on decelerating learning continuously due to net force on it acting down the plane. If a be the deceleration of the block, then the linear deceleration of the rim of wheel is also equal to a but angular deceleration of the wheel is D = a/r..

Now, if T is the tension in the sting, then equations of motion are :

Mg sin T - T = Ma ………. (1) And Tr = ID = I (a/r) ………. (2) Eliminating T, we get 2 2 2 sin sin , 1 I a Mgr Mg Ma or a r Mr T T 

Initial velocity of the block up the plane = v = Zr Thus distance moved by the block before stopping is

2 2 2 2 2 2 2 2 2 sin 2 sin r I Mr I Mr v s a M gr Mg Z Z T T   .

1. A right circular cylinder of mass m and radius r is suspended from a cold that is wound around its circumference. If the cylinder is allowed to fall.

Kinetics of rigid body: Force and acceleration:

Let the acceleration of the mass centre G be a and its angular accelerationD . Equations of motion of the cylinder can be written as, x x F ma 6 : No forces acting : y y F ma 6 mg – T = ma ……….. (i) : G G G M I D Tr I D 6 2 2 m r Tr D ……….. (ii) As there is no slip, , a a r r D D .

Eliminating T from (i) and (ii) and substituting for D

2 mr a mg ma r r   2 a ga

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2 3 a g _Ans: From (i) 2 3 T m ga m g§_¨  g·_¸ © ¹ 3 mg T Ans.

Problem 11: The moment of inertia of a reel of thread of radius R and mass M about its axis is I. If the loose and of the thread is held in hand and rest is allowed to unroll itself falling down under the action of gravity show that.

a. It falls down with an acceleration, 2

1 /

g a

I MR 

Solution: (a) When the reel unrolls itself under the action of gravity, the downward force due to gravity Mg, must exceed the upward force i.e., tension T of the thread. Hence, if a is the tangential acceleration of a point on the rim of reel, Newton’s second law of motion yields

Mg – T = Ma ………… (1)

The resultant torque on the reel is TR, so that from W ID we get

TR = ID or TR = Ia/R ……… (2)

Where a and D represent linear and angular accelerations respectively.. Eliminating T from (1) and (2), we have

M (g – a) R = Ia/R or a = g/[1 + (I/MR2_{)] ……… (3)}

30.14 Motion of Two Bodies Connected By a String and Passing Over a Pulley:

Consider two bodies connected by an inextensible light string and passing over a simple pulley as shown in figure. 30.10

Let m₁ and m₂ = Masses of the two bodies in kg (such that their weights are m₁.g and m₂.g in Newton’s)

Let m₁ be greater than m₂. A little consideration will show that the greater mass m₁ will move downwards, whereas the smaller one will move upwards.

Let a = Acceleration of the two bodies, P₁ and P₂ = Pulls in the string in Newton’s

D = Angular acceleration of the pulley,, r = Radius of the pulley,

M = Mass of the pulley,

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I = Moment of inertia of the pulley, and k = Radius of gyration of the pulley.

First of all, consider the motion of body 1 of mass m₁ kg, which is coming down. We know that the forces acting on it are m₁ g Newton’s (downwards) and P₁ Newton’s (upwards). As the body is moving downwards, therefore, resultant force

= m₁.g – P Newton’s ………….. (i)

Since the body is moving downwards with an acceleration (a) therefore force acting on it.

= m₁.a ………….. (ii)

Equating equations (i) and (ii),

m₁.g – P = m₁ a. ………….. (iii)

Now consider the motion of body 2 of mass m₂ kg, which is going upwards. We know that the forces acting on it are m₂.g Newton’s (downwards) and P₂ Newton’s (upwards). As the body in moving upwards, therefore, resultant force

= P₂ – m₂.g Newton’s

Since the body is moving upwards with an acceleration (a) therefore force acting on it

= m₂.a ………….. (v)

Equating equations (iv) and (v). P₂ – m₂.g = m₂. a

Now consider motion of the pulley, which is rotating about its axis due to downward motion of the body of mass m₁ tied to the string. We know that linear acceleration of the body 1 is equal to the angular acceleration of the pulley.

.

a rD

? ………….. (vii)

And torque, T = Net tension in the string u Radius of pulley

1 2

PP r ………….. (viii)

We also know that torque on the pulley,

.

T ID ………….. (ix)

Equating equations (viii) and (ix),

1 2 .

P P r ID

2 1 2 . .

P P r IDr (Multiplying both sides by r)

= I. a …….. ? a r x. 2 ₂ . I a P P r ? 

Problem 42: In the arrangement shown in figure 59, the mass of the uniform solid cylinder of radius R is equal to m and the masses of the two bodies are equal to m₁ and m₂. The thread slipping and the friction in the axle of the cylinder are supposed to be absent. Calculate the angular acceleration of the cylinder and the ratio of tensions T₁/T₂ of the vertical sections of the thread in the process of motion.

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Solution: The free – body – diagram is shown in figure 60. Because the pulley has got a mass hence tensions in the two strings will be different. Hence, equation of motion for mass m₁ :

m₁ g – T₁ = m₁a ………. (1)

For mass m₂ : T₂ = m₂g = m₂a ………. (2) For rotational motion of cylinder : (T₁ – T₂) R = ID ………. (2)

Here a represents linear acceleration and Dimplies angular acceleration Adding equations (1) and (2), m₁g – T₁ + T₂ - m₂ g = (m₁ + m₂)a. Or m1m2 g T1T2 m1m2 DR Or m1m2 g ID/R m1m2 DR [From eq. (3)] Or m1m2 g ª_¬m1m2 I R/ 2 º_¼DR 1 2 2 1 2 / m m g R m m I R D  ? ª _{ } º ¬ ¼ 1 2 1 2 2 2 1 2 1 2 2 2 2 / 2 / m m g m m g R m m m R m m I m R R     ª _{ } º ¬ ¼

From equation (2) and (3), we get : 1 2 2 2 2 2 2 1 [ ( / ] T I I I T R T R m a m g R m g R D D D D    2 1 2 1 2 2 [1 ( / )] [1 { (2 2 ) / 2( ) } I I g m m m R m m g R m R  g RD     2 1 2 1 2 2 2 1 2 1 2 1 2 (1 / 2) 2 ( ) ( ) [4 ] [2 2 2 2 ] mR m m m m m m m m m R m m m m m u        1 1 2 2 2 2 1 2 1 ( ) (4 ) 1 (4 ) ( ) T m m m m m m T m m m m m m   ?    .

Problem 43: In the system shown in figure 61, the masses of the two bodies are m₁ and m₂, the coefficient of friction between the body m₁ and the horizontal plane is equal to P, and a pulley of mass m is assumed to be a uniform disc. The thread does not slip over the pulley. At the moment t = 0 the body m₂ starts descending. Assuming

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the mass of the thread and the friction in the axle of the pulley to be negligible, calculate the work performed by the frictional forces acting on the body m₁ over the first t second after the beginning of the motion.

Solution: Free body diagram is shown in figure 62. If a and D represent the linear and angular acceleration respectively,, then equations of motion for linear and rotational motions are given by:

1 1 1 T Pm g m a ……… (1) 2 2 2 m g T m a ……… (2) 2 1 T T R ID ……… (3)

Adding equation (1) and (2), we get

2 1 2 1 1 2 T T m gPm g m m a Or 2 1 1 2 2 1 1 2 2 ( ) / ( ) 2 ( ) m m g m g m g I R m m a or a m m m P P D       2 1 [ / ( ) ] 2 a R and I m R D 

Suppose the Mass m₂ falls through h, hence

2 2 1 2 1 2 1 2 1 2 2 2 1 1 2 2 2 2 m m gt m m h at and v at gt m m m m m m P P   u    

According to law of convention of energy : 2 1 12 1 22 2

1 1 1 2 2 2 m gh m v  m v  IZ W Also, v1 v2 v ZR at 2 2 2 2 2 2 2 1 2 2 1 1 1 / 2 / 2 2 2 f m gh m a t m a m R a t R W ?   u  Or 4m gh2 2m22m1m a t2 2Wf 2 2 1 2 2 2 2 1 1 2 4 2 2 2( f m m gt W m g m m m a t m m m P  ?     

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2 2 2 2 2 2 1 2 1 2 1 2 1 2 4 4 2 2 m m g t m m g t m m m m m m m P P        2 2 2 1 2 1 1 2 1 1 2 2 1 2 1 2 4 4 2 2 m m m m m g t m m m g t m m m m m m P P P P ª    º  « »     « » ¬ ¼ .

Kinetics of Motion of Rotation Dividing both sides by 2,

147 – P2 = 0.95 D ……….. (x)

From equation (vi) we find that

P₂ – 98 = 10 a₂ = 10 u 0.1 D = D ……….. (xi) Adding equations (x) and (xi),

49 = 1.95 ……….. (x).

2

49 / 1.95 25.1rad s/ D

?

Pulls in the two parts of the string

Substituting the value of D in equation (x).

2

147P 0.95D 0.95 25.1u 23.8

2 147 238 123.2

P N

?  Ans.

Now substituting the value of D and P₂ in equation (ix),

1 0.4 25.1 123.2 133.2

P u  N Ans:

Example 30.14 Two bodies A and B of mass 150 kg and 75 kg respectively are supported by a string of negligible mass and pass over a composite pulley. The bodies rest on two smooth inclined planes as shown in figure 30.13.

If the pulley has a mass of 75 kg and radius of gyration of 100 mm, find the accelerations of the masses A and B pulls in the strings. Naglet friction in the bearings.

Solution: Given; m₁ = 150 kg; m₂ = 75 kg; M = 75 kg; k = 100 mm = 0.1 m; r₁ = 250 mm = 0.25 m; r₂ = 125 mm = 0.125 m.

Pulls in the strings

Let P₁ = Pull in the string carrying 150 kg mass, and P₂ = Pull in the string carrying 75 kg mass.

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From the geometry of the masses, we find that the turning moment of mass 150 kg (i.e., 150 sin 450 _{u 0.25 = 150 u} 0.707 u 0.25 = 26.5 kg – m). It is thus obvious that the 150 kg mass will come downwards and the 75 kg mass will go upwards, when the system is released.

Example 30.13: Two masses of 30 kg and 10 kg are tied to the two ends of a light string passing over a

composite pulley of radius of gyrations as 70 mm and mass 4 kg as shown in figure 30.12 Find the pulls in the two parts of the string and the angular acceleration of the pulley.

Solution: Given: m₁ = 30 kg; m₂ = 10 kg; k = 70 mm = 0.07 m; M == 4 kg; d₁ = 100 mm = 0.1 m or r₁ = 0.05 m; Angular acceleration of the pulley

Let P₁ = Pull in the string carrying 30 kg mass, P₂ = Pull in string carrying 10 kg mass, and

D = Angular acceleration of the body..

From the geometry of the masses, we find that turning moment of the mass 30 kg (i.e., 30 u 0.05 = 1.5 kg – m) is more than that of the mass 10 kg (i.e., 10 u 0.1 = 1 kg – m). It is thus obvious, that the 30 kg mass will come downwards and the 10 kg mass will go upwards, when the system is released.

Substituting the value of P₁ in equation (iii)

1039 – (0.5 P₂ + 3D ) = 150 a₁ = 150 u 0.25 D 1039 – 0.5 P₂ - 3D = 37.5 D

2

1039 0.5P 37.5D 3D 40.5D

?  

Multiplying both sides by 2 …………. (x)

2078 – P₂ = 81 D

From equation (vi), we find that

2 367.5 75 2 75 0.125 9.4

P  a u D D

Adding equation (x) and (xi) 1710.5 = 90.4 D

2

1710.5 / 90.4 18.9rad s/ D

?

Now substituting the value of D in equation (x),

2

2078P 81D 81 18.9u 1531

2 2078 1531 547

P N

?  Ans:

Again substituting the value of D and P₂ in equation (ix),

1 0.5 547 3 18.9 330

P u  u N Ans.

Acceleration of the masses A and B

We know that the acceleration of mass A (i.e., 150 kg)

2 1 r1. 0.25 18.9 4.72m s/

D D u

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Similarly, 2 2 r2. 0.125 18.9 2.36m s/

D D u

SAMPLE PROBLEMS 16.2

The thin plate ABCD of mass 8 kg is held in the position shown by the wire BH and two links AE and DF. Neglecting the mass of the links, determine immediately after wire BH has been cut (a) the acceleration of the plate, (b) the force in each link.

SOLUTION:

Kinematics of Motion: After wire BH has been cut, we observe that corners A and D move along parallel circles of radius 150 mm centered, respectively, at E and F. The motion of the plate is thus a curvilinear translation; the particles forming the plate move along parallel circles of radius 150 mm.

At the instant wire BH is cut, the velocity of the plate is zero. Thus the acceleration _a& of the mass center G of the plate is tangent to the circular path which will be described by G.

Equations of Motion: The external forces consist of the weight W and the forces F_AE, and F_DF exerted by the links. Since the plate is in translation, the effective farces reduce to the vector m a

&

attached at G and directed along the t axis. A free-body-diagram equation is drawn to show that the system of the external forces is equivalent to the system of the effective forces.

a. Acceleration of the Plate. t _{t eff} F F  6 6 0 cos 30 W ma 0 cos 30 mg ma 0 2 0 cos 30 9.81 / cos 30 a g m s 2 0 8.50 / 60 a m s  …. (1)

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30 W G 200 mm F F 30 30 250 mm n t A B C D

b. Forces in Links AE and DF.

0 : sin 30 n n eff AE DF F F F F W  6 6   : G _{G eff} M M  6 6 _{…. (2).} 0 0 sin 30 250 cos 30 100 AE AE F m m  F mm 0 0 sin 30 250 cos 30 100 0 DF DF F mm F mm   38.4FAE211.6FDF 0 0.1815 DF AE F  F …. (3).

Substituting for F_DF from (3) into (2), we write

0 0.1815 sin 30 0 AE AE F  F W F_AE = 0.6109 W 0.1815 0.6109 0.1109 DF F  W  W

Noting that W = mg = (8 kg) (9.81 m/s2_{) = 78.48 N, we have} F_AE = 0.6109 (78.48 N) F_AE = 47.9 N T

F_DF = - 0.1109 (78.48 N) F_DF = 8.70 N C.

Example 22.9: A thin uniform bar of mass m and length L is suspended from two vertical inextensible strings. If the right hand string BD is cut find the tension in the left string AC and the angular acceleration of the bar (Fig. 22.10).

Solution: As the string BD is cut, the bar AB starts rotating about the end A. It should be noted here that the bar does

not rotate about its mass centre G.

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Let the angular acceleration of the bar be D then the linear acceleration of the mass centre G is

2 L D§ ·_{¨ ¸}

© ¹ and its

direction would be normal to the bar at G. Forces acting on the bar are shown in the figure. 22.10.

The equations of motion can be written as

x x F ma 6 : No force x y F ma 6 _{: mg – T = ma ……… (i)} : 2 G G G L M I D T§ ·I D 6 _{¨ ¸} © ¹ As, 2 12 G m L I Therefore, 2 2 12 m L T L _D ……… (ii)

Eliminating T from (i) and (ii) and using

2 L

a D .

KINETICS OF RIGID BODY: FORCE AND ACCELERATION:

2 2 2 2 m L L L L mg ma mg mD D § · _ § _ · ¨ ¸ ¨ ¸ © ¹ © ¹ 3 2 g L D 2 3 2 / 2 12 2 G I m L g T L L L D § · ¨ ¸ © ¹ 4 m g T Ans:

Example 22.10: A thin uniform rod AB of length L = 1 m and mass m = 10 kg is hinged at the point C which is at a distance of 0.25 m from the end A (Fig. 22.11). The rod is released from the horizontal position. Find (a) the angular velocity of the rod when it has rotated through 300 _{(b) the reaction at the hinge.}

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Solution: The bar AB when released rotates about the hinge C. The mass centre G rotates in a circle of radius L/4 (0.25 m). When the bar is at an angle T = 300 _{from the horizontal let its angular velocity be Z angular} acceleration be D . Component of the linear acceleration of the mass centre G, along the bar

2 4 n L a Z § ·_{¨ ¸} © ¹

Component of the linear acceleration of the mass centre G, normal to the bar

4

t

L a ¨ ¸D§ ·

© ¹

Forces acting on the bar are (i) the weight of the bar mg (ii) reaction of the hinge on the bar having components R_n and R_t as shown in figure 22.11.

Equation of motion of the bar can written as

: cos 4 ( ) t t t t L F ma mg R ma m

Normal to the bar

D T § · 6  _{¨ ¸} © ¹ _{……….. (i)} 2 : sin 4 ( ) n n n t L F ma R mg ma m

Along the bar

Z T § · 6  ¨_¨ ¸_¸ © ¹ _{……….. (ii)} 4 12 G G t G L m L M I D R § · I D § ·D 6 _{¨ ¸ ¨ ¸} © ¹ © ¹ ……….. (iii)

Eliminating R_t from (i) and (iii) and solving for D

2 4 cos 4 12 L m L m mg L D D T § ·_{¨ ¸} © ¹ cos 4 3 L L g D §_¨  ·_¸ T © ¹ 12 cos 7 g D T 16.82 cos D T When 0 30 , 14.56rad s/ , T D . d d d d d t d d t d Z Z T Z D Z T T

Equation (iv) can be written as

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16.82 cos T Z d_dZ_T 16.82 cos d d Z Z T T. Integrating from 00 _{to 30}0 0 30 ₂ 0 0 Re 2 16.82 cos tan lation cannot d d g

be used as is not cons t Z Z D T Z Z T T D § _ · ¨ ¸ ¨ ¸ © ¹

³

³

To find the components of the reaction R_t and R_n, substitute of ZandD in the equations (i) and (ii)

0 10 14.65 cos 10 9.81 cos 30 4 4 t m L R mg T D u u  u u

Kinetics of rigid body: force and acceleration:

R_t = 48.5 N 2 2 0 10 4.10 1 sin 10 9.81 sin 30 4 4 n m L R Z mg T u u  u R_n = 91.1 N.

Reaction at the hinge 2 2 2 2

48.5 91.1

t n

R R R 

R = 103.2 N Ans.

Angular Acceleration Produced by a torque:

In terms of moment of inertia, equation (7.18) can be written as net ID

* ……….(7.19 a).

The vector form of the above equation is net ID

* )& )))&

……….(7.19 b).

The angular acceleration produced is along the direction of applied net external torque. The magnitude of the produced angular velocity acceleration is directly proportional to the magnitude of the net torque and inversely proportional to the moment of inertia of the body. The above relation looks the translational equation _F)&_{net ma}&. Here, you should not forget that * ID is not an independent. It is derived from F = ma only. We can establish and analogue between translational and rotational variables. By doing so concept developed so far for translational motion would help to solve the problems involving rotational motion. The possible analogue is as follows.

Translational Variable/Expression Rotational Variable/Expression (i) distance traversed, S Angle turned, T

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(ii) average speed, v '_'s_t Average angular speed, w '_{d t}T

(iii) instantaneous speed, v ds

dt Average angular acceleration, t Z

D '

'

(iv) average acceleration, a d v_{d t} Average angular acceleration, D '_{d t}Z

(v) instantaneous acceleration, a d v_{d t} Instantaneous angular acceleration, D d_{d t}Z (vi) mass, m Moment of inertia I

(vii) force, F Torque, *

(viii) F = ma * ID

(ix) linear momentum, p Angular momentum, l (x) p = mv l IZ

(xi) F d p_{d t} * d l_{d t}

(xii) conversation of linear momentum : When F = 0, p = constant. Conservation of angular momentum:When

0, l

* constant.

(xiii) translational kinetic energy, 1 2

2

k mv Rotational kinetic energy 1 2

2

k IZ

(xiv) work done, dZ F ds. Work done, dZ *.dT

Angular quantities involved in analogues (ix) to (xiv) would be discussed later in this topic.

Example 11:

Find the angular acceleration of the rod given in example 4 at the moment.

a. When it is released from rest in the horizontal position; b. When it makes an angle T with the horizontal.

Solution: After the moment when the rod is released from the rest in the horizontal position, it would rotate in the vertical plane about a horizontal axis passing through the hinge and perpendicular the length of the rod. Initial angular velocity of the rod is zero but due to non zero torque of gravity it has some angular acceleration and hence, it will acquire some angular speed as it rotates. As discussed in example 4, the hinge force does not provide any torque about the axis under consideration and the weight of the rod tries to rotate it in the clockwise sense, i.e., it provides a torque perpendicular, inward to the plane of the paper. An approach using _{r F}&_u)&to find the torque would also give the same result. Hence, angular acceleration of the rod,

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gravity net I I D * * 2 . / 2 / 3 mg l ml 3 2 g l .

(b) When the rod makes an angle T with the horizontal, its angular acceleration.

mg net I I D * * . mg r I A 2 . .cos 2 2 l mg ml T [from figure 7.40 (b)] = 3 .cos 2 g l T. Example 12:

In the previous case, find the angular velocity of the rod when it has turned through an angle T after the moment when it was released from rest in the horizontal position. Also find the angular velocity when the rod becomes vertical.

Solution: From the result obtained in part (b) of the previous example, at some angleT, the angular acceleration of the rod, 3 .cos 2 g l D T 3 cos 2 d g d t l Z T Ÿ

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3 . cos 2 d d g d t d t l Z T _T

Ÿ _{[Using chain rule.]}

0 3 . cos . 2 g d d l Z Z Z T T Ÿ

³

When the rod becomes vertical, T S/ 2, and hence, angular velocity,, 3 /g l

Z .

Example 13: In the previous example, find the hinge force on the rod at T 0.

Solution: Just after the moment when the released from the rest in the horizontal position, it is shown in figure 7.14 (a). Let the vertical component of the force on the rod from the hinge be R₁ and the horizontal component of the same be R₂, as shown in figure. The subsequent motion of the center of mass of the rod is a nouniform circular notice on the vertical circular path of radius l/2 with the centre at the hinge, as suggested in the figure.

Initially the rod is at rest hence radial component of the acceleration of the centre of mass of the rod, _Z2_{r, is zero.}

Hence, applying Fnet macm along the radial direction, we get

, ,

net radial cm radial

F ma

2 0

R m

Ÿ u

= 0

Applying the same along the tangential direction, we have,

, tan ,tan

net gential cm gential

F ma 1 2 l mg R m§D · Ÿ  _{¨ ¸} © ¹ 1 2 l R mg mD Ÿ 

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3 4

mg mg

1 4mg

Net hinge force 12 22

1 4

R R mg. [R2 0]

Example 14: In the previous example, find the magnitude of the net hinge force on the when the rod has turned through an angle T.

Solution: If Z be the angular velocity of the rod when it has turned through an angle T, the centre of mass of the rod has 2

2 2

l l

and

Z D as radial and tangential components of its acceleration, respectively, as shown in figure 7.41 (b). Applying Fext Macm on the rod along the radial direction, we have,

2 2 sin 2 l R mg T mZ 2 3 sin sin 2 R mg T mg T Ÿ  3 sing l T Z 5 sin 2mg T. 3 cos 2 g l D T

Applying the same along the tangential direction, we have,

1 cos . 2 l mg TR mD 1 cos 2 l R mg T mD Ÿ  3 cos cos 4 mg T mg T 1 cos 4mg T

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Therefore, net force on the rod from the hinge can be obtained by solving

2 2 1 2

R R R .

Example 15: If the disc given in example 5 has mass M and it is free to rotated about its symmetrical axis passing through O, find its angular acceleration.

Solution: If D be the angular acceleration of the disc, using *net ID , we have,

net I D * 2 3 / 2 F R MR 6 F MR.

As the net torque is in clockwise sense, D has the same sense of rotation.

Example 16: A uniform disc of radius 0.l2 m and mass 5 kg is pivoted so that it rotates freely about its axis. A thin, mass less and inextensible sting wrapped around the disc is pulled with a force of 20 N, as shown in figure 7.42(a). (a) What is the torque exerted on the disc about its axis?

(b) What is the angular acceleration of the disc?

(c) If the disc starts from rest, what is the angular velocity after 3 s?

Solution: It is obvious that the string force gives a torque to the disc in the clockwise direction. As the torque given by the force from the axle is zero. Net torque on the disc is,

net

* = torque of the string force = F r._A

= F. R

= (20. N). (0.12 m) = 2.4 N – m.

As the net torque on the disc is in clockwise direction, the disc has angular acceleration in the same direction. If Dbe the magnitude of the angular acceleration,

2 / 2 net net I _MR D * * 2 2 2.4 2 / 5 0.12 rad s u u 2 66.66rad s/  .

At t = 0 if the disc has zero angular velocity, then, at some time t, its angular velocity, in t

Z Z D

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t Z D Ÿ

? At t = 3 s,

Z = 200 rad/s.

Example 17: A uniform disc of radius R and mass M is mounted on an axis supported in fixed frictionless bearing. A light string is wrapped around the rim of the disc and a body of mass m is supported by the strings, as shown in figure 7.43 (a).

a. Find the angular acceleration of the disc;

b. Find the magnitude of the tangential acceleration of the point on the rim where the string separates from the rim.

c. If the system is released from rest at t = 0, find the speed of the block at some time t (> 0).

Solution: Analyze the situation according to the information provided in the figure 7.43 (b). You should also note the following points:

· Only tension force of the siring, T, produces a torque on the disc about its centre 0. Torque of the weight of the disc and that of the reaction force from the bearing are zero about 0.

· If D be the angular acceleration of the disc (in the clockwise direction) then the point P on the disc has a tangential acceleration D R in the vertically downward direction at the moment shown in figure. The string unwinds at the same acceleration and the block has the same acceleration in the vertically downward direction. Therefore, if a be the acceleration of the block, then,

a = DR ………… (i)

Now, applying *net ID on the disc about its symmetrical axis, we have, net ID * 2 . . 2 M R T R D Ÿ 2 M R T D Ÿ 2 Ma T

Ÿ [Using (i) …………... (ii)

Using Fnet = ma for the block in the vertical direction, we have, net

F = ma

mg T ma

Ÿ  …………... (iii)

Adding (ii) and (iii), we get,

2 M mg §_¨ m a·_¸ © ¹ . / / 2 m a g R m M § · Ÿ _{¨ ¸}  © ¹ .

If v be the speed of the block at some time t, then, we have, v = u + at [ a is constan ]t

= at [u 0]

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. / 2 m g t m M § · ¨ _ ¸ © ¹

Example 18: Find the acceleration of m₁ and m₂ in an Atwood’s Machine, shown in figure 7.44 (a), if there is friction present between the surface of pulley and the thread does not slip over the surface of the pulley. Moment of inertia of the pulley about its symmetrical axis is I and its radius is R. The pulley can rotate freely about its symmetrical axis.

Solution: Due to friction between the pulley and the thread tensions in the parts of the thread on the two sides of the pulley are different. Let that in the right part it is T₁ and that in the left part is T₂, as shown in figure 7.44 (b).Forces acting on the two blocks and the pulley are also shown in figure 7.44(b). Force on the pulley from the support and its weight are not shown because they do not produce torque on the pulley about its symmetrical axis of rotation. If the block m₁ comes down with an acceleration a then m₂ would go up with the same acceleration because they are connected by the same string, as shown in the same figure.7.44(b).

If we assume that the pulley gets an angular acceleration in the clockwise sense then the torque of T₁ would be positive and that of T₂ would be negative, as suggested in figure 7.44(c).

Again, as any point on the rim of the pulley has a tangential acceleration DR, the block m₁ comes down and the block m₂ goes w with the same acceleration, as shown in figure 7.44(d).

Therefore, we can write,

A = D R ……….. (i)

Using Fnet = ma for the two blocks, we have,

m₁ g – T₁ = m₁ a ……….. (ii) [For m₁] T₂ – m₂ g = m₂ a ……….. (iii) [For m₂]

Using *net ID for the pulley, we have,

+ T₁. R – T₂. R = ID ……….. (iv) [ torque of support force and

weight are zero] Substituting from (i) in (iv), we get,

1 2 ₂

a

T T I

R

 _{……….. (v)}

Adding (ii), (iii) and (v), we get,

1 2 1 2 2 I m m g m m a R § ·  _¨   _¸ © ¹ 1 2 2 1 2 / m m a g m m I R §  · Ÿ ¨_¨ ¸_¸   © ¹

Example 19: A thin uniform rod AB of mass m = 1.0kg moves translationally with acceleration a = 2.0 rn/s2 _{due to} two anti parallel force F₁ and F₂ acting on it perpendicular to its length, as shown in figure 7.45. The distance between the points at which these forces are applied is x = 20 cm. Besides, it is known that F₂ = 5.0 N. Find the length of the rod.

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Solution: Before analyzing the details of the given situation, let us analyze the rotational effect of two anti parallel forces. Consider the situations shown in figure 7.45.

If we analyze the torques of the two forces about every point in their plane containing them, then, we arrive at the conclusion that if the point lies between the lines of action of F₁ and F₂ then torques of the forces about that point add up together otherwise they are in opposite directions.

If the magnitudes of the two forces are equal then such a pair is called as a couple. If the magnitude of each force is F and the distance between their lines of application is d, then, the net torque about any point in their plane is F. d, asshowninfigure7.47.

O

Now, let us discuss the given case. As the rod is in pure translation motion, net torque on it about any point must be zero. Therefore, the centre of mass of rod can not lie between the lines of action of the forces because in

that case torques produced by then about the centre of mass do not cancel each other.

Let us assume that the centre of mass of the rod lies at a distance y away from the line of action of F₂ as shown in figure 4.48. As the rod translates towards right, F2 must have a greater magnitude thanF₁.

Using F_net = ma, we have F₂ – F₁ = ma 1 2 F F ma Ÿ 5 1 2 N u = 3 N.

Again, as the net torque on the rod about C must be zero,

1 2

magnitude of the torque magnitude of the torque produced by F about C produced by F about C

the two torques have opposite directions ? ª º « » ¬ ¼ 1 2 F x y F y Ÿ 

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