Question

A solid cylinder of radius 25cm is released from rest at the top of a smooth 4degree incline of height h=5m. At the same time and from the same position, a hollow sphere of radius 25 cm is also released from res. Both objects roll without slipping and both objects have a mass of 75g.

How long did it take for the faster of the two objects to reach the bottom of the hill?

Answer 1

angle, theta=4 degrees

radius, r=25cm

height, h=5m

use,

in case of solid cylinder,

m*g*h=1/2*I*w^2+1/2*m*v^2

m*g*h=1/2*(1/2)m*r^2*w^2+1/2*m*v^2

g*h=1/4*r^2*(V^2/r^2)+1/2*v^2

g*h=1/4*(V^2)+1/2*v^2

g*h=3/4*v^2

===> v=sqrt(4*g*h/3)

v=sqrt(4*9.8*5/3)

v=8.08 m/sec

time taken to reach the bottom,

t=h*sin(theta)/v

t=5*sin (4)/8.08

t=43.2 msec ------------->

and

in case of hollow sphere,

m*g*h=1/2*I*w^2+1/2*m*v^2

m*g*h=1/2*(2/3)m*r^2*w^2+1/2*m*v^2

g*h=1/3*r^2*(V^2/r^2)+1/2*v^2

g*h=5/6*(V^2)+1/2*v^2

g*h=5/6*v^2

===> v=sqrt(6*g*h/5)

v=sqrt(6*9.8*5/5)

v=7.67 m/sec

time taken to reach the bottom,

t=h*sin(theta)/v

t=5*sin (4)/7.67

t=45.5 msec --------------->