Physics - Rotation Question with Solution | TestHub
Match the following statements related to pure rolling motion with their correct descriptions or expressions.
List - I | List - II |
|---|---|
(P) Condition for pure rolling of a wheel of radius R | (1) |
(Q) Total kinetic energy of a body of mass M and radius R undergoing pure rolling (K is radius of gyration) | (2) |
(R) Velocity of the point of contact with the ground for a body in pure rolling | (3) |
(S) Velocity of the topmost point of a wheel of radius R undergoing pure rolling, if its center of mass moves with speed | (4) |
Options:
Answer:
Solution:
P. For pure rolling, the condition is that the velocity of the center of mass is equal to .
Q. The total kinetic energy of a body undergoing pure rolling is the sum of its translational and rotational kinetic energies: .
R. In pure rolling, the point of contact with the ground is instantaneously at rest, so its velocity is .
S. The velocity of the topmost point of a wheel in pure rolling is the sum of the center of mass velocity and the tangential velocity due to rotation: . P -> (2): v_CM = R omega. Q -> (4): K = (1/2)Mv_CM^2 + (1/2)I_CM omega^2. R -> (1): the contact point has zero velocity. S -> (3): v_top = 2v_CM. Therefore the matching code is option A.
