Answer to Question #87356 in Mechanics | Relativity for Sridhar

Question

a ring and a disc of same mass roll without slipping along a horizontal surface with same velocity. if the kinetic energy of ring is 8J then that of disc is

ans:6J

Answer 1

The ring and the disc which are rolling without slipping take part both the translational and rotational movements simultaneously, therefore their full kinetic energy is

$$W=\frac{M{{v}^{2}}}{2}+\frac{J{{\omega }^{2}}}{2}$$

where

$$\frac{M{{v}^{2}}}{2}={{W}_{T}}$$

is the Translational Kinetic Energy

$$\frac{J{{\omega }^{2}}}{2}={{W}_{R}}$$

is the Rotational Kinetic Energy, M is the mass, v is the velocity, J is the moment of inertia, ω is the angular velocity.

The moments of inertia of the ring and the disc with radii R_r and R_d are

$$J=MR_{r}^{2},\,\,J=\frac{MR_{d}^{2}}{2}$$

respectively.

Then the Kinetic Energy of the ring W_r is

$${{W}_{r}}=\frac{M{{v}^{2}}}{2}+\frac{MR_{r}^{2}\omega _{r}^{2}}{2}$$

When rotating without slipping we have

$${{R}_{r}}{{\omega }_{r}}={{v}_{r}}=v$$

Then

$${{W}_{r}}=\frac{M{{v}^{2}}}{2}+\frac{M{{v}^{2}}}{2}=M{{v}^{2}}=8J$$

The Kinetic Energy of the disc W_d is

$${{W}_{r}}=\frac{M{{v}^{2}}}{2}+\frac{MR_{d}^{2}\omega _{d}^{2}}{4}=\frac{M{{v}^{2}}}{2}+\frac{M{{v}^{2}}}{4}=\frac{3M{{v}^{2}}}{4}$$

Since

$$M{{v}^{2}}=8J$$

then

$${{W}_{r}}=\frac{3M{{v}^{2}}}{4}=\frac{3}{4}\cdot 8=6J$$

Thus, the Kinetic Energy of disc is 6J.