Question

19. cm In figure E and V represent the total energy and speed of centre of mass of an object of mass 1 kg in pure rolling. The object is: E(J) 3 Vem(m? S-2) 0 4 (A) sphere (C) disc (B) ring (D) Hollow Cylinder​

Answer 1

Answer:

Whole body is in pure rotation wrt pt of contact.

V=Rω

V

T

=2Rω

Adding the linear velocity with rotational velocity vectorially,we get,

Velocity at top is 2v and,

Velocity at bottom point is zero as it is the point of no slipping.

Answer 2

Your question was incomplete. Please check below the full content.

Diagram has been attached

Concept:

• Centre of mass of rolling objects
• The kinetic energy of rolling objects
• Rotational motion
• Moment of Inertia

Given:

• E/V^2 = 3/4

Find:

• The exact object whose graph has been depicted

Solution:

Energy E = 1/2mv^2 [1 + k^2/R^2]

E/v^2 =  1/2m [1 + k^2/R^2]

m = 1 kg

E/v^2 =  1/2 [1 + k^2/R^2]

3/4 = 1/2 [1 + k^2/R^2]

3/2 =  1 + k^2/R^2

1/2 = k^2/R^2

For a disc, I = mR^2/2 = mk^2

mR^2/2 = mk^2

1/2 = k^2/R^2

The correct option is C) disc.

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