Question

A solid sphere, solid cylinder, and a hollow pipe all have equal masses and radii and are of uniform density. If the three are released simultaneously at the top of an inclined plane and roll without slipping, which one will reach the bottom first? A. solid sphere B. hollow pipe C. solid cylinder D. They all reach the bottom at the same time

Answer 1

Answer:

A )  SOLID SPHERE

Explanation:

Moment of inertia of solid sphere = 2/5 M R²

= M K² , K is called radius of gyration

K = √2/5 R

Moment of inertia of solid cylinder = 1/2  M R²

= M K² , K is called radius of gyration

K = 1 /√2  R

Moment of inertia of solid sphere =  M R²

= M K² , K is called radius of gyration

K =  R

For rolling on inclined plane , acceleration  

a = $\frac{gsin\theta}{1+\frac{K^2}{R^2} }$

Putting the value of K for solid sphere

a for solid sphere

a = g sinθ / ( 1 +2/5)

a = .714 g sinθ

Putting the value of K for solid cylinder

a for solid cylinder

a = g sinθ / ( 1 +1/2)

a = .666 g sinθ

Putting the value of K for hollow pipe

a for hollow pipe

a = g sinθ / ( 1 +1 )

a = . 5 g sinθ

So we see that acceleration a for solid sphere is greatest and a for hollow pipe is the least. Hence solid sphere will reach the bottom earliest and hollow pipe will reach the bottom the latest.