Question

A solid sphere, solid cylinder, and a hollow pipe all have equal masses and radii. If the three of them are released simultaneously at the top of an inclined plane and do not slip, which one will reach the bottom first?
(A) sphere
(B) pipe
(C) cylinder
(D) The pipe and cylinder arrive together before the sphere.
(E) They all reach the bottom at the same time.

Answer 1

Answer:

The solid sphere will reach the bottom first.

Explanation:

In order to develop this problem and give it a correct solution, it is necessary to collect the concepts related to energy conservation. To apply this concept, we first highlight the importance of conserving energy so we will match the final and initial energies. Once this value has been obtained, we will concentrate on finding the speed, and solving what is related to the Inertia.

In this way we know that,

$\Delta KE = - \Delta PE$

$KE_t + KE_r = mgh$

We know as well that the lineal and angular energy are given by,

$KE_r = \frac{1}{2}I\omega^2$

And the tangential kinetic energy as

$KE_t = \frac{1}{2} mv^2$

Where$\omega = \frac{v}{R}$

Replacing

$\frac{1}{2}mv^2 + \frac{1}{2}I\frac{v}{R} = mgh$

Re-arrange for v,

$v=\sqrt{\frac{2mgh}{m+I/R^2}}$

We have here three different objects: solid cylinder, hollow pipe and solid sphere. We need the moment inertia of this objects and replace in the previous equation found, then,

For hollow pipe:

$I_{hp}=mR^2$

$v_{hp}=\sqrt{\frac{2mgh}{m+(mR^2)/R^2}}$

$v_{hp}=\sqrt{\frac{2mgh}{m+m)}$

$v_{hp}=\sqrt{gh}$

For solid cylinder:

$I_{sc}=\frac{1}{2}mR^2$

$v_{sc}=\sqrt{\frac{2mgh}{m+(1/2mR^2)/R^2}}$

$v_{sc}=\sqrt{\frac{2mgh}{m+1/2m}}$

$v_{sc}=\sqrt{\frac{3}{4}gh}$

For solid sphere,

$I_{ss}=\frac{2}{5}mR^2$

$v_{ss}=\sqrt{\frac{2mgh}{m+(2/5mR^2)/R^2}}$

$v_{ss}=\sqrt{\frac{2mgh}{m+2/5m}}$

$v_{ss}=\sqrt{\frac{10}{7}gh}$

Then comparing the speed of the three objects we have:

$v_{hp}$

$\sqrt{gh}$