A solid disc is rolling without slipping at constant speed on a horizontal surface. What is the ratio of its rotational kinetic energy to its translational kinetic energy

2/7

Explanation:

The Rotational Kinetic Energy is given as RKE = 1/2 * I * w², where

w = v/R and

I = 2/5mR²

The Translational Kinetic Energy is given as TKE = 1/2 * m * v²

RKE = 1/2 * 2/5mR² * (v/R)²

RKE = 1/2 * 2/5mR² * v²/R²

RKE = 1/5mv²

To get the ratio, we use the relation

RKE / (TKE + RKE)

and thus, we have

1/5mv² / (1/2mv² + 1/5mv²)

1/5mv² / 7/10mv²

1/5 ÷ 7/10

1/5 * 10/7

2/7

Therefore, the ratio of RKE to TKE is 2/7

i guess it’s a and b

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