
A cylinder of mass and radius is rolling without slipping on a horizontal surface with angular velocity . The velocity of the center of mass of the cylinder is . The cylinder comes across a step of height . Then the angular velocity of the cylinder just after the collision is (Assume cylinder remains in contact and no slipping occurs on the edge of the step.)
(A)
(B)
(C)
(D)

Answer
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Hint Hint: To solve this question, we need to use the theorem of conservation of angular momentum. For that, we have to equate the angular momentum of the cylinder about the edge of the step, before and after the collision.
Formula used: The formulae used to solve this question are
(A) , here is the angular momentum, is the angular velocity , and is the moment of inertia.
(B) , here is the angular momentum of an object about a point moving with a velocity of at a perpendicular distance of from the point.
Complete step-by-step solution:
The rolling cylinder will interact with the step at the point P as shown in the below figure.
As we can see that a normal force acts on the cylinder at the point P. Since passes through the point P, so its torque about the point P is equal to zero. This means that the angular momentum of the cylinder about P is conserved.
Initial angular momentum:
In the above figure, the cylinder is rotating with the angular velocity of and it is also translating with a velocity of about the point P. So the total initial angular momentum about P is given by
.........(1)
The perpendicular distance of the velocity vector from the point P from the above figure can be given by
..................(2)
Also, we know that the moment of inertia of a cylinder about its natural axis is given by
…………………..(3)
Putting (2) and (3) in (1) we get
On simplifying, we get
……..(4)
Final angular momentum:
Let the final angular velocity of the cylinder be , as shown in the below figure.
Since it is given that no slipping occurs on the edge, the translational velocity of the centre of mass of the cylinder is given by
…………..(5)
Since the cylinder is translating with velocity and rotating about the point P with angular velocity , so its final angular momentum about P is given by
In this case, the perpendicular distance of P from the velocity vector is equal to the radius, as can be seen in the above figure. So we can put in the above equation to get
Putting (3) and (5) in the above equation, we get
On simplifying we get
…………..(6)
Now, by the conservation of angular momentum, we equate (4) and (6) to get
Cancelling from both the sides, we get
Multiplying with we finally get
Hence, the correct answer is option A.
Note: We can also attempt this question using the concept of instantaneous axis of rotation. The given cylinder will have a pure rotation about the instantaneous axis passing through P and perpendicular to the plane of the cylinder. So its angular momentum about P will be given by , where will be the moment of inertia about the instantaneous axis of rotation, which can be obtained by using the parallel axis theorem.
Formula used: The formulae used to solve this question are
(A)
(B)
Complete step-by-step solution:
The rolling cylinder will interact with the step at the point P as shown in the below figure.

As we can see that a normal force
Initial angular momentum:
In the above figure, the cylinder is rotating with the angular velocity of
The perpendicular distance of the velocity vector from the point P from the above figure can be given by
Also, we know that the moment of inertia of a cylinder about its natural axis is given by
Putting (2) and (3) in (1) we get
On simplifying, we get
Final angular momentum:
Let the final angular velocity of the cylinder be

Since it is given that no slipping occurs on the edge, the translational velocity of the centre of mass of the cylinder is given by
Since the cylinder is translating with velocity
In this case, the perpendicular distance of P from the velocity vector is equal to the radius, as can be seen in the above figure. So we can put
Putting (3) and (5) in the above equation, we get
On simplifying we get
Now, by the conservation of angular momentum, we equate (4) and (6) to get
Cancelling
Multiplying with
Hence, the correct answer is option A.
Note: We can also attempt this question using the concept of instantaneous axis of rotation. The given cylinder will have a pure rotation about the instantaneous axis passing through P and perpendicular to the plane of the cylinder. So its angular momentum about P will be given by
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