Problem 1. A particle of mass m is projected at time t = 0 from a point P with a speed v at an angle of 45º to the horizontal. Find the magnitude and the direction of the angular momentum of the particle about the point P at time t.
0
Solution: Let us take the origin at P, x-axis along the horizontal and y-axis along the vertically upward direction as shown in figure.
For horizontal motion during the time 0 to t,
0 0
For vertical motion,
0 0
The angular momentum of the particle at time t about the origin is
(
ˆ ˆ) (
x ˆ y ˆ)
Thus, the angular momentum of the particle is 0 2 2 2
mv gt in the negative z-direction i.e., perpendicular to the plane of motion, going into the plane.
Problem 2. A small solid sphere of mass 1 kg and radius 0.2 metre rolls down along a track shown in figure, without slipping. Find the height h above the base, from which it has to start rolling down the incline, such that the sphere just completes the vertical circular loop of radius 1 metre.
A r h R
Solution: The situation is shown in figure. Let m and r be the mass and radius of solid sphere. When the sphere reaches the point A , it descends through a vertical distance
( )
, where isthe radius of circular loop. At , the loss of potential energy
2
h− R r+ R
A
=mg h
(
−2R r+)
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Physics : Rotational Motion
Total kinetic energy of the sphere at A
= K.E. of the rotation + K.E. of translation of sphere
2 2
For just completing the vertical circular loop
( )
mv2 mg R r =
−
or v2 =
(
R−r)
g .Now total kinetic energy of sphere at 7 2 A =10mv 7
( )
10m R r g
= − .
Loss of potential energy = gain of K.E.
(
2)
7( )
h− R r+ =10m R r g−
mg .
Solving we get,
27 17
10 R 10
= × −
h r
substituting the given values, we get
27 17
Problem 3. Calculate the kinetic energy of a tractor crawler belt (see figure) of mass m if the tractor moves with velocity v. There is no slipping. The
dimensions of the wheels are as shown in the figure. v
A B
C D
Solution: The velocity of the centre of mass of the tractor wheel is v . velocity of the lower part of the belt in contact with the ground is
= −v v =0.
Velocity of the upper part between the two wheels is = + =v v 2v
Suppose that the mass of the upper portion of the belt AB between the two wheels is m1 and the total mass of the portion of the belt in contact with the wheels between BC and DA is m2. The total mass of the belt is:
(
2m1+m2)
.Physics : Rotational Motion
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The kinetic energy (KE) of the belt
= KE of AB + KE of CD + KE of BC + KE of DA
Problem 4. A cylinder of mass m is suspended through two strings wrapped around it as shown in figure. Find (a) the tension T in the string and
(b) the speed of the cylinder as it falls through a distance h. T T
mg
Solution: The portion of the strings between the ceiling and the cylinder is at rest. Hence the points of the cylinder where the strings leave it are at rest. The cylinder is thus rolling without slipping on the strings. Suppose the centre of the cylinder falls with an acceleration a. The angular acceleration of the cylinder about its axis is a
α = , as the cylinder does not slip over the r strings.
The equation of motion for the centre of mass of the cylinder is
mg −2T =ma … (i)
and for the motion about the centre of mass, it is
1 2 1
As the centre of the cylinder starts moving from rest, the velocity after it has fallen through a distance h is given by
2 2 connected to a link and freely rotates about with angular velocity 60 radian/second. If the rotor is suddenly allowed to rest on a vertical wall, what time will elapse before it comes to rest?
The coefficient of friction between the wall and the rotor is 0.25 and inclination of link with respect to the vertical is 15º.
W
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Physics : Rotational Motion
Solution: In the figure, is the normal reaction at C and T is the tension in AB . The rotor has no motion whether horizontal or vertical.
R
Angular acceleration, torque
Moment of inertia Time taken by the rotor to come to rest
ω =ω0+αt or, 0 60 5.92t= − t =10.13 sec.
Problem 6. A round cone with half-angle α = 30º and the radius of the base R = 10 cm rolls uniformly without slipping over a horizontal plane. The apex is hinged at the point O (see figure) which is at the same level as the point C, the centre of the base. The velocity of the point C is . Find (a) the angular velocity vector of the cone and the angle it forms with the vertical and (b) the angular acceleration vector.
v 10ms= −1 C
O
Solution: (a) There are two angular motions of the cone; one in the horizontal plane with angular velocity 1
cot v
R α
ω = where = radius of the circle in which C moves. The direction of this vector is upward. The other angular velocity is about its own axis.
Since it rolls without slipping, cot horizontal and towards O .
= ω and hence, the angular acceleration corresponding to is zero. There is no change in the magnitude of but there is change in its direction. ∆ θ= angle through which the axis OC rotates in the horizontal plane .
ω1
Physics : Rotational Motion
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change in angular velocity
Problem 7. Two horizontal discs rotate freely about a common vertical axis passing through their centres. The moments of inertia of the discs relative to there axis are equal to and , and their angular velocities are and . The upper disc falls on the lower disc and after some time, both discs begin to rotate as a single body due to friction. Find
I1 I2 ω1 ω2
(a) the steady state angular velocity of the discs, and (b) the work performed by the frictional forces in the process.
Solution: (a) Since there is no external torque on the system consisting of both the discs, angular momentum of the system is conserved.
I1 1ω + ω =I2 2
(
I1+I2)
ω ⇒f 1 1 2 2(b) Work done by frictional forces = change in kinetic energy of the system Wfriction= change in kinetic energy =Tf −Ti
which is very similar to the expression for a perfectly inelastic collision.
Problem 8. A uniform sphere of mass m and radius R starts rolling without slipping down an inclined plane. Find the time dependence of the angular momentum of the sphere relative to the initial point of contact. How will the result be affected in the case of a perfectly smooth inclined plane? Here, θ = angle of inclination of the plane.
Solution: Let O be the initial point of contact.
For linear motion , we write, mgsinθ −ffr =macm.
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Physics : Rotational Motion
or, n I2 cm I2 dvcm
Using the expression for angular momentum
(
Laxis=Lspin+Lorbital)
0 cm If the plane is perfectly smooth, there is no rotation.
∴ L0 =mv Rcm =mgsinθRt .
Problem 9. A truck, initially at rest with a solid cylindrical paper roll, moves forward with a constant acceleration a. The cylinder roll is lying parallel to the forward edge of the truck at a distance d from the rear edge of the truck. Find the distance s which the truck travels before the paper roll moves off the edge of its horizontal surface. Friction is sufficient to prevent slipping between the paper and the truck.
d
Solution: Let a be the acceleration of the cylinder relative to truck.
The different forces acting on the cylinder are shown in fig. Here,
where t is the time required to cover the distance d.
t 3d a
∴ =
The distance traveled by the truck relative to ground before the paper rolls off is given by
1 2 1 3
Physics : Rotational Motion
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Problem 10. A uniform sphere of mass m and radius r rolls without slipping down from the top of a larger fixed sphere of radius R. Find the angular momentum acquired by smaller the sphere about the centre of the larger sphere as well as its K.E. at the moment when it breaks off from the larger sphere. The initial velocity of the sphere is negligible.
Solution : The centre of mass of the sphere moves in a vertical circle of radius
(
R r+)
.Considering the dynamics of circular motion of the centre of mass of the sphere
mv2
At the point where the smaller sphere breaks off, m 2 2r
Since friction is static and gravity is conservative (other forces being constraint forces), mechanical energy is conserved.
( ) ( )
cos 1 2 1 2The angular momentum of the sphere is L L= orbital+ ωI
Kinetic energy of the sphere is
K.E. 1 2 1 2
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Physics : Rotational Motion