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

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

Tardigrade Admin · Accepted Answer

Total kinetic energy of the ring when it rolls without
slipping,

K_{ring} = K_{T} + K_{R} = \frac{1}{2} m v^{2} + \frac{1}{2} I_{r} ω^{2}

= \frac{1}{2} m v^{2} + \frac{1}{2} m r^{2} \times \frac{v ^{2}}{r ^{2}} = m v^{2} (∵ I_{r} = m r^{2}

and

ω = \frac{v}{r})

But

K_{ring} = 4 J

(given)

∴ m v^{2} = 4 J ... (i)

Similarly,

K_{disc} = \frac{1}{2} m v^{2} + \frac{1}{2} I_{d} ω^{2}

= \frac{1}{2} m v^{2} + \frac{1}{2} \times \frac{m r ^{2}}{2} \times \frac{v ^{2}}{r ^{2}} (∵ I_{d} = \frac{m r ^{2}}{2})

= \frac{3}{4} m v^{2} = \frac{3}{4} \times 4 J = 3 J

Using (i))

Q. 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 4J then total kinetic energy of the disc is

3 J

4 J

5 J

6 J

Q. 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