Question

A hoop, a uniform solid cylinder, a spherical shell, and a uniform solid sphere are released from rest at the top of an incline. What is the order in which they arrive at the bottom of the incline?

Answer 1

Explanation:

Given

Hoop, Uniform Solid Cylinder, Spherical shell and a uniform Solid sphere released from Rest from same height

Suppose they have same mass and radius

time Period is given by

`t=\sqrt{(2h)/(a)}` ,where h=height of release

a=acceleration

`a=(g\sin \theta )/(1+(I)/(mr^2))`

Where I=moment of inertia

a for hoop

`a=(g\sin \theta )/(1+(mr^2)/(mr^2))`

`a=(g\sin \theta )/(2)`

a for Uniform solid cylinder

`a=(g\sin \theta )/(1+(mr^2)/(2mr^2))`

`a=(2g\sin \theta )/(3)`

a for spherical shell

`a=(g\sin \theta )/(1+(2mr^2)/(3mr^2))`

`a=(3g\sin \theta )/(5)`

a for Uniform Solid

`a=(g\sin \theta )/(1+(2mr^2)/(5mr^2))`

`a=(5g\sin \theta )/(7)`

time taken will be inversely proportional to the square root of acceleration

`t_1=k√(2)=1.414k`

`t_2=k\sqrt{(3)/(2)}=1.224k`

`t_3=k\sqrt{(5)/(3)}=1.2909k`

`t_4=k\sqrt{(7)/(5)}=1.183k`

thus first one to reach is Solid Sphere

second is Uniform solid cylinder

third is Spherical Shell

Fourth is hoop