Question

A ring and a disc roll on the horizontal surface without slipping with same linear velocity. If both have same mass and total kinetic energy of the ring is $4\, J$ then total kinetic energy of the disc is

• A. 3 J
• B. 4 J
• C. 5 J
• D. 6 J

Answer 1

Answer: (A)

Total kinetic energy of the ring when it rolls without
slipping, $K_{\text {ring }}=K_{T}+K_{R}=\frac{1}{2} m v^{2}+\frac{1}{2} I_{r} \omega^{2}$
$=\frac{1}{2} m v^{2}+\frac{1}{2} m r^{2} \times \frac{v^{2}}{r^{2}}=m v^{2} \,\,\,\,\left(\because I_{r}=m r^{2}\right.$ and $\left.\omega=\frac{v}{r}\right)$
But $K_{\text {ring }}=4 J$ (given) $\therefore m v^{2}=4 J \,\,\,\, ...(i)$
Similarly, $K_{\text {disc }}=\frac{1}{2} m v^{2}+\frac{1}{2} I_{d} \omega^{2}$
$=\frac{1}{2} m v^{2}+\frac{1}{2} \times \frac{m r^{2}}{2} \times \frac{v^{2}}{r^{2}} \quad\left(\because I_{d}=\frac{m r^{2}}{2}\right)$
$=\frac{3}{4} m v^{2}=\frac{3}{4} \times 4 J =3 J \,\,\,\,$ Using (i))