Question

Two uniform solid balls, one of radius R and mass M, the other of radius 2 R and mass 8 M, roll down a high incline. They start together from rest at the top of the incline. Which one will reach the bottom of the incline first?

Answer 1

Answer:

First ball reaches at the bottom first.

Explanation:

m1 = M

r1 = R

m2 = 8M

r2 = 2R

Moment of inertia of the solid ball is

I = 2/5 mr^2

So, the radius of gyration of the ball

$K=\sqrt{\frac{2}{5}}r$

So,

$K_{1}=\sqrt{\frac{2}{5}}R$

$K_{2}=\sqrt{\frac{2}{5}}\times 2R$

The ball having more acceleration reaches the first at the bottom of the inclined plane.

The acceleration is given by

$a=\frac{gSin\theta }{1+\frac{K^{2}}{r^{2}}}$

So,

$a_{1}=\frac{gSin\theta }{1+\frac{2}{5}}=\frac{5gSin\theta }{7}$

$a_{2}=\frac{gSin\theta }{1+\frac{8}{5}}=\frac{5gSin\theta }{13}$

So, a1 > a2

Thus, the first ball reaches at the bottom first.