Circular Motion :
When a particle moves on a circular path with uniform speed, its is said to execute a uniform circular motion.
Angular Velocity : It is the rate of change of angular displacements of the body. If the radial line in the adjoining figure rotates through an angle θ(radian) in time t (seconds) then its angular velocity.
O θ
ω = t
θ radian / second
If it takes the radial line a time T to complete one revolution, then
ω = T 2π
and if n revolutions are made in 1s then n = T
1 and ω = 2πn
The angular acceleration of the particle is given by α = ∆t
ω
∆
Linear Velocity :
Linear velocity = angular velocity × radius v = ω × r
linear acceleration of particle (a) = a × r
Centripetal Acceleration : When a particle moves with uniform speed v in a circle of radius r it is acted upon by an acceleration v2/r in the direction of centre.
It is called centripetal acceleration. The acceleration has a fixed magnitude but its direction is continuously changing. It is always directed towards the centre of the circle.
Centripetal Forces : If the particle of mass m moves with uniform velocity v in circle of radius r, then force acting on it towards the centre is
r
mv2 . This is called centripetal force. It has a fixed magnitude and is always directed towards the centre.
Without centripetal force, a body can not move on a circular path. Earth gets this force from the gravitational attraction between earth and sun;
electron moves in circular path due to electrostatic attraction between it and nucleus. A cyclic or car while taking turn, gets the centripetal force from the friction between road and type. To create this force, the vehicle tilts itself towards the centre. If it makes angle θ with the vertical in tilted position then than θ = v2/rg. where v is its velocity and r is the radius of the path. In order to avoid skidding (or slipping), the angle of tilt θ with vertical should be less than angle of friction λ. i.e. tan θ < tan λ
or rg
v2 < µ (since coefficient of friction µ = tan λ)
In limiting condition rg
v2 = µ or v = µ.r.g
This is the maximum safe speed at the turn.
Since centripetal force is provided by the friction, it can never be more than the maximum value µR = (µmg) or frictional force.
Motion in a vertical circle : When a body tied at one end of a string is revolved in a vertical circle, it has different speed at different points of the circular path.
Therefore, the centripetal force and tension in the string change continuously. At the highest point A of motion.
mg Ta
r
Tb
vb
B mg va A
C
Ta + mg = r
mv2a or Ta = r mv2a
– mg This tension, at highest point will be zero, for a minimum velocity vc given by
0 =
r mv2c
– mg or vc = gr
Circular Motion, Rotational Motion
P HYSICS F UNDAMENTAL F OR IIT-J EE
KEY CONCEPTS & PROBLEM SOLVING STRATEGY
This minimum speed is called critical speed (vc). If the speed at A is less than this value, the particle will not reach up to the highest point. To reach with this speed at A, the body should have speed at B given by the conservation laws viz.
Decrease in kinetic energy = increase in potential energy
Therefore, the body should have speed at B at least gr
5 , so that it can just move in vertical circle.
Tension in string at B is given by.
Tb – mg =
This means that the string should be able to stand to a tension, equal to six times the weight of the body otherwise the string will break.
At any other point P making angle θ with the vertical, from the figure.
A
Conical pendulum :
A conical pendulum consists of a string AB (fig.) whose upper end is fixed at A and other and B is tied with a bob. When the bob is drawn aside and is given a horizontal push. Let it describe a horizontal circle with constant angular speed ω in such a way that AB makes a constant angle θ with the vertical. As the string traces the surface of a cone, it is known as conic pendulum.
Let l be the length of string AB. The forces acting on the bob are (i) weight mg acting downwards, (ii) tension T along the sting (horizontal) component is T sin θ and vertical component is T cos θ).
T cos θ = mg
The horizontal component is equal to the centripetal force i.e.,
Rotational Motion :
Centre of mass of a system of particles :
The point at which the whole mass of the body may be supposed to be concentrated is called the centre of mass.
Consider the case of a body of an arbitrary shape of n XY plane as shown in fig. Let the body consist of number of the coordinates of centre of mass, then
x =
When there is a continuous distribution of mass instead of being discrete, we treat an infinitesimal element of the body of mass dm whose position is (x, y, z). In such a case, we replace summation by integration in above equations. Now we have,
x =
z =
Consider two particles of masses m1 and m2 located at position vectors r1 and r2 respectively with respect to origin. Now the position vector r of the centre of mass is given by
(m1 + m2)r = m1r1 + m2r2 ...(1) Thus, the product of the total mass of the system and position vector of the centre of mass is equal to the sum of the products of the individual masses and their respective position vectors. Hence
r =
Now the velocity of centre of mass of the system is given by v =
dt dr
The acceleration of the centre of mass is given by
a =
The equation describing the motion of the centre of mass may be written as
f(total) = M dt dv
When no external force acts on the system, then 0 = M
Therefore, when no external force acts on the system, the centre of mass of an isolated system move with uniform velocity.
Moment of inertia and radius of gyration :
Moment of Inertia : The moment of inertia of a body about an axis is defined as the sum of the products of the masses of the particles constituting the body and the square of their respective distance from the axis.
Radius of Gyration : If we consider that the whole mass of the body is concentrated at a distance K from the axis of rotation, then moment of inertia I can be expressed as
I = MK2
where M is the total mass of the body and K is the radius of gyration. Thus the quantity whose square when multiplied by the total mass of the body gives the moment of inertia of the body about that axis is known as radius of gyration.
Theorems on moment of inertia :
Theorem of parallel axes : According to this theorem, the moment of inertia I of a body about any axis is equal to its moment of inertia about a parallel axis through centre of mass IG plus Ma2 where M is the mass of the body and a is the perpendicular distance between the axes, i.e., I = IG + Ma2
Theorem of perpendicular axes : According to this theorem, the moment of inertia I of the body about a perpendicular axis is equal to the sum of moment of inertia of the body about two axes right angles to each other in the plane of the body and intersecting at a point where the perpendicular axis passes, i.e.,
I = Ix + Iy
Table of moment of inertia :
Body Axis Moment of
inertia 1. Thin uniform rod
of length l
Through its centre and perpendicular to its length
12 Ml2
2. Thin rectangular sheet of sides a and b.
Through its centre and perpendicular to its plane
rectangular bar of length l, breadth b and thickness t.
Through its midpoint and perpendicular to its length sphere of radius R
About a diameter 5 2MR2
5. Circular ring of
radius R. Through its
centre and perpendicular to its plane
MR2
6. Disc of radius R. Through its centre and perpendicular to its plane centre and parallel to its length
(ii) Through its centre and perpendicular to its length.
2
Angular momentum of a rotating body :
In case of rotating body about an axis, the sum of the momentum of the linear momentum of all the
particles about the axis of rotation is called angular momentum about the axis.
Q Also the angular momentum of rigid body about an axis is the product of moment of inertia and the angular velocity of the body about that axis.
L = r × p = Iω
Translational and rotational quantities :
Translational Motion Rotational Motion Displacement = s Angular displacement = θ Velocity = v Angular velocity = ω Acceleration = a Angular acceleration = α Inertia = m Moment of inertia = I
Force = F Torque = τ
Momentum = mv Angular momentum = Iω Power = Fv Rotational power = τω Kinetic energy =
2
1mv2 Rotational K.E. = 2 1Iω2
Kinematics equation of a rotating rigid body :
The angular velocity of a rotating rigid body is defined as the rate of change of angular displacement, i.e., →ω= (d→θ/dt)
Similarly, the angular acceleration is defined as the rate of change of angular velocity, i.e.,
→α = dt d→ω
= 22 dt d →θ
Let a body be rotating with constant angular acceleration →α with initial angular velocity ω→0. If θ is the initial angular displacement, then its angular velocity →ω and angular displacement θ at any time is given by the following equations
ω = ω0 + αt 0 = ω0t +
21 αt2 and ω2 = ω02 + 2 αθ
These equations are similar to usual kinematics equation of translatory motion.
v = u + at, s = ut + 2 1at2 and v2 = u2 + 2as
Problem Solving Strategy : Rotational Dynamics for Rigid Bodies :
Our strategy for solving problems in rotational dynamics is very similar to the strategy for solving problems that in involve Newton’s second law.
Step-1 : Identify the relevant concepts : The equation Στ = Iαz is useful whenever torques act on a rigid body–that is, whenever forces act on a rigid body in such a way as to change the state of the body’s rotation.
In some cases you may be able to use an energy approach instead. However, if the target variable is a force, a torque, an acceleration, an angular acceleration, or an elapsed time, the approach using Στ = Iα2 is almost always the most efficient one.
Step-2 : Setup the problem using the following steps:
Draw a sketch of the situation and select the body or bodies to be analyzed.
For each body, draw a free-body diagram isolating the body and including all the forces (and only those forces) that act on the body, including its weight. Label unknown quantities with algebraic symbols. A new consideration is that you must show the shape of the body accurately, including all dimensions and angles you will need for torque calculations.
Choose coordinate axes for each body and indicate a positive sense of rotation for each rotating body. If there is a linear acceleration, it’s usually simplest to pick a positive axis in its direction. If you know the sense of αz in advance, picking it as the positive sense of rotation simplifies the calculations. When you represent a force in terms of its components, cross out the original force to avoid including it twice.
Step-3 : Execute the solution as follows :
For each body in the problem, decide whether it under goes translational motion, rotational motion, or both. Depending on the behavior of the body in question, apply ΣF = m ar, Στz = Iαz, or both to the body. Be careful to write separate equations of motion for each body.
There may be geometrical relations between the motions of two or more bodies, as with a string that unwinds from a pulley while turning it or a wheel that rolls without slipping. Express these relations in algebraic form, usually as relations between two linear accelerations or between a linear acceleration and an angular acceleration.
Check that the number of equations matches the number of unknown quantities. Then solve the equations to find the target variable(s).
Step-4 : Evaluate your answer : Check that the algebraic signs of your results make sense. As an example, suppose the problem is about a spool of thread. If you are pulling thread off the spool, your answers should not tell you that the spool is turning in the direction the results for special cases or intuitive expectations. Ask yourself : Does this result make sense ?”
Problem Solving Strategy: Equilibrium of a Rigid Body Step-1 : Identify the relevant concepts : The first and
second conditions for equilibrium are useful whenever there is a rigid body that is not rotating and not accelerating in space.
Step-2 : Set up the problem using the following steps:
Draw a sketch of the physical situation, including dimensions, and select the body in equilibrium to be analyzed.
Draw a free-body diagram showing the forces acting on the selected body and no others. Do not include forces exerted by this body on other bodies. Be careful to show correctly the point at which each force acts; this is crucial for correct torque calculations. You can't represent a rigid body as a point.
Choose coordinate axes and specify a positive sense of rotation for torques. Represent forces in terms of their components with respect to the axes you have chosen; when you do this, cross out the original force so that you don't included it twice.
In choosing a point to compute torques, note that if a force has a line of action that goes through a particular point, the torque of the force with respect to that point is zero. You can often eliminate unknown forces or components from the torque equation by a clever choice of point for your calculation. The body doesn't actually have to be pivoted about an axis through the chosen point.
Step-3 : Execute the solution as follows :
Write equations expressing the equilibrium conditions. Remember that ΣFx = 0, ΣFy = 0, and Στz = 0 are always separate equations; never add x-and y-components in a single equation. Also remember that when a force is represented in term of its components, you can compute the torque of that force by finding the torque of each component separately, each with its appropriate lever arm and sign, and adding the results. This is often easier than determining the lever arm of the original force.
You always need as many equations as you have unknowns. Depending on the number of unknowns, you may need to compute torques with respect to two or more axes to obtain enough equations. Often, there are several equally good sets of force and torque equations for a particular problem; there is usually no single "right"
combination of equations. When you have as many independent equations as unknowns, you can solve the equations simultaneously.
Step-4 : Evaluate your answer : A useful way to check your results is to rewrite the second condition for equilibrium, Στz = 0, using a different choice of origin. If you've done everything correctly, you'll get the same answers using this new choice of origin as you did with your original choice
Solved Examples
1. A particle a moves along a circle of radius R = 50 cm so that its radius vector r relative to point O (fig.) rotates with the
A
R O C
r
A
R O C
r Y
θ X θ
(a) (b)
constant angular velocity ω = 0.40 rad/sec. Find the modulus of the velocity of the particle and modulus and direction of its total acceleration.
Sol. Consider X and Y axes as shown in fig. Using sine law in triangle CAO, we get
) 2 sin(
r θ
−
π =
sinθ R or
θ θcos sin 2
r =
sinθ R
∴ r = 2 R cos θ
Now r = r cos θ i + r sin θ j
= 2 R cos2θ i + 2R cos θ sin θ j
Now, v
dt
dr = – 4R cos θ sin θ dt dθ
i + 2R cos 2θ dt dθ
j
= – 2 R sin 2 θ ω i + 2 R cos 2θ ω j
∴ |v| = 2 ω R Further
a = dt
dv = 4 R cos 2 θ dt dθ
i – 4 R ω sin 2 θ dt dθ
j = –4 R ω2 cos 2 θ i – 4R ω2 sin 2θ j
|a| = 4 Rω2
2. A particle describes a horizontal circle on the smooth inner surface of a conical funnel as shown in fig. If the height of the plane of the circle above the vertex 9.8 mark cm, find the speed of the particle.
Sol. The forces acting on the particle are shown in fig.
They are
R sin α α R
r h=9.8 cm α mg mv2/r
R cos α
(i) weight m g acting vertically downwards.
(ii) normal reaction R of smooth surface of the cone.
(iii) reaction of the centripetal force (mv2/r) acting radially outward.
Hence, R sin α = m g ...(1) and R cos α = (mv2/r) ...(2)
Dividing eq. (1) by eq. (2), we get tan α =
v2
r
g ...(3)
From figure, tan α = (r/h) ...(4) From eqs. (3) and (4), we get
h r = 2
v r
g or v = (gh)
∴ v = [9.8 × (9.8 × 10–2)]1/2 = 9.98 m/s
3. A particle of 10 kg mass is moving in a circle of 4m radius with a constant speed of 5m/sec. What is its angular momentum about (i) the centre of circle (ii) a point on the axis of the circle and 3 m distant from its centre ? Which of these will always be in same direction ?
Sol. The situation is shown in fig.
L1
L2
3 0 4
5
(a) We know that L = r × mv L = m v r sin θ
Here m = 10 kg, r = 4 m, v = 5 m/sec and θ = 90º
∴ L = 10 × 5 × 4 × 1 = 200 kg-m2/sec.
(b) In this case r = (42+32) = 5m.
∴ L = 10 × 5 × 5 = 250 kg-m2/sec.
From figure it is obvious that angular momentum in first case always has same direction but in second case the direction changes.
4. A symmetrical body is rotating about its axis of symmetry, its moment of inertia about the axis of rotation being 1 kg-m2 and its rate of rotation 2 rev./sec. (a) what is its angular momentum ? (b) what additional work will have to be done to double its rate of rotation ?
Sol. (a) As the body is rotating about its axis of symmetry, the angular momentum vector coincides with the axis of rotation.
∴ Angular momentum L = Iω ...(1)
Kinetic energy of rotation E = 2 1Iω2 or 2E = Iω2
∴ I2ω2 = 2IE or Iω = (2IE) ...(2) From eqs. (1) and (2), L = (2IE) ...(3) ω = 2 rev/sec = 2 × 2π or 4π radian/sec.
∴ E = 2
1 × 1 × (4π)2 = 8π2 joule
Now L = (2IE) = (2×1×8π2) = (16π2)= 4π
= 12.57 kg.m2/sec.
(b) When the rate of rotation is doubled, i.e., 4 rev/sec or 8π radians/sec, the kinetic energy of rotation is given by
E = 2
1 × 1 × (8π)2 = 32π2 joule Additional work required
= Final K.E. of rotation – Initial K.E. of rotation = 32π2 – 8π2
= 24 π2 = 236.8 joule
5. A thin horizontal uniform rod AB of mass m and length l can rotate freely about a vertical axis passing through its end A. At a certain moment the end B starts experiencing a constant force F which is always perpendicular to the original position of the stationary rod and directed in the horizontal plane. Find the angular velocity of the rod as a function of its rotation angle φ counted relative to the initial position.
Sol. The situation of the rod at an angle φ is shown in fig.
Here
r = i l cos φ + + j l sin φ and F = j F
(Force is always perpendicular to rod)
A B X
F Y F
lθ
→τ = r × F = (i l cos φ + j l sin φ) × (j F) = l F cos φ k
| →τ| = l F cos φ We know that τ = 1 α Here I =
3
1m l2 (for rod) and α = ω (dω/dφ)
∴ l F cos φ = 3
1 m l2 . ω (dω/dφ)
or l F cos φ dφ = 3
1 m l2 . ω dω Integrating within proper limits, we have
l F
∫
0φcosφdφ = 31m l2
∫
0ωωd ω l F sin φ = 31m l2(ω2/2) ∴ ω =
φ ml
sin F 6
Addition of hydrogen halides to Alkenes : Markovnikov’s Rule
Hydrogen halides (HI, HBr, HCl, and HF) add to the double bond of alkenes :
C = C + HX → – C – C – H H
These additions are sometimes carried out by dissolving the hydrogen halide in a solvent, such as acetic acid or CH2Cl2, or by bubbling the gaseous hydrogen halide directly into the alkene and using the alkene itself as the solvent. HF is prepared as polyhydrogen fluoride in pyridine. The order of reactivity of the hydrogen halides is HI > HBr > HCl
> HF, and unless the alkene is highly substituted, HCl reacts so slowly that the reaction is not one that is useful as a preparative method. HBr adds readily, the reaction may follow an alternate course. However, adding silica gel or alumina to the mixture of the alkene and HCl or HBr in CH2Cl2 increases the rate of addition dramatically and makes the reaction an easy one to carry out.
The addition of HX to an unsymmetrical alkene could conceivably occur in two ways. In practice, however, one product usually predominates. The addition of HBr to propene, for example, could conceivably lead to either 1-bromopropane or 2-bromopropane. The main product, however is 2-bromopropane :
CH2 = CHCH3 + HBr → CH3CHCH3
Br 2-Bromopropane
When 2-methylpropene reacts with HBr, the main product is tert-butyl bromide, not isobutyl bromide :
C = CH2 + HBr → CH3 – C – CH3
Br tert-Butyl bromide
CH3
H3C H3C
2-Methylpropene (isobutylene)
Consideration of many examples like this led the Russian chemist Vladimir Markovnikov in 1870 to formulate what is now known as Markovnikov’s rule. One way to state this rule is to say that in the addition of HX to an alkene, the hydrogen atom adds to the carbon atom of the double bond that
already has the greater number of hydrogen atoms. The addition of HBr to propene is an illustration :
already has the greater number of hydrogen atoms. The addition of HBr to propene is an illustration :