Question

A coin and a ring have the same mass and radius. Starting from rest and the same height, they roll down the same incline. In other words, they both start with the same gravitational potential energy. Which one of the following statements is true about the coin and ring when they reach the bottom of the incline?

A. The increase in the coin's total kinetic energy is less than the increase in the ring's total kinetic energy.
B. The increase in the coin's total kinetic energy is equal to the increase in the ring's total kinetic energy.
C. The increase in the coin's total kinetic energy is greater than the increase in the ring's total kinetic energy. A. Coin’s Krotational = ring's Krotational,and coin's KCM = ring's KCM B. Coin’s Krotational = ring's Krotational,and coin's KCM < ring's KCM C. Coin’s Krotational = ring's Krotational,and coin's KCM > ring's KCM D. Coin’s Krotational < ring's Krotational,and coin's KCM < ring's KCM E. Coin’s Krotational < ring's Krotational,and coin's KCM = ring's KCM F. Coin’s Krotational < ring's Krotational,and coin's KCM > ring's KCM G. Coin’s Krotational > ring's Krotational,and coin's KCM > ring's KCM H. Coin’s Krotational > ring's Krotational,and coin's KCM = ring's KCM I. Coin’s Krotational > ring's Krotational,and coin's KCM < ring's KCM A. The speed of the coin's center of mass is less than the speed of the ring's center of mass. B. The speed of the coin's center of mass is equal to the speed of the ring's center of mass. C. The speed of the coin's center of mass is greater than the speed of the ring's center of mass.

Answer 1

Coin is a disc, so its moment of Inertia I = mr²/2

I of ring = mr²

Now, by conservation of mechanical energy,

Decrease in Potential energy (U)= Rotational Kinetic Energy(K_rot) + Translational Kinetic Energy (K_tra)

K_rot = 1/2*I*w² where w is the angular velocity

K_tra = 1/2 mv² where v is the velocity of the centre of mass

Applying conservation of energy for disc

mgh = 1/2 * mr²/2 * w² + 1/2 * m * v²

also, for rolling v = wr, so substituting w = v/r in the above equation

mgh = 1/2 * mr²/2 * (v/r)² + 1/2 * m * v²

Solving we get v = (4gh/3)^1/2

K_tra = 1/2 * m * v² = 2mgh/3

K_rot = 1/2 * mr²/2 * (v/r)² = mgh/3

Applying conservation of energy for ring

mgh = 1/2 * mr² * w² + 1/2 * m * v²

also, for rolling v = wr, so substituting w = v/r in the above equation

mgh = 1/2 * mr² * (v/r)² + 1/2 * m * v²

Solving we get v = (gh)^1/2

K_tra = 1/2 * m * v² = mgh/2

K_rot = 1/2 * mr² * (v/r)² = mgh/2

By conservation of energy potential energy is converted to total kinetic energy (roational + translational). As same amout of potential energy is lost same amout of kinetic energy is gained by both the bodies. So option B

From above calculations, K_rot of coin< K_rot of ring and K_tra of coin > K_tra​ of ring, so option F

v of coin is (4gh/3)^1/2 and v of ring is (gh)^1/2 so option C