Question

Compared to an object that does not roll, but instead slides without friction, should a rolling object be released from the same,a greater, or a lesser height in order just barely to complete the loop the loop? view available hint(s) compared to an object that does not roll, but instead slides without friction, should a rolling object be released from the same,a greater, or a lesser height in order just barely to complete the loop the loop? the rolling object should be released from a greater height. the rolling object should be released from a lesser height. the rolling object should be released from exactly the same height.

Answer 1

The correct answer is:
 - the rolling object should be released from a greater height

In fact:
- for the sliding object, the initial mechanical energy of the object is the gravitational potential energy:
E=mgh
where h is the height at which it is released. At the top of the loop, this energy has converted into sum of potential energy and kinetic energy:
$E= \frac{1}{2}mv^2 + mg(2R)$
where $v= \sqrt{gR}$ is the minimum velocity the object should have at the top of the loop, and R is the radius of the loop.
Since energy must be conserved, we can write:
$mgh=\frac{1}{2}mv^2 + mg(2R)$
from which the height at which the object should be released is
$h= \frac{1}{2} \frac{v^2}{g} +2R$

- for the rolling object, we can do the same considerations, however at the top of the loop the object has also rotational energy:
$E= \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2+ mg(2R)$
where I is the moment of inertia of the object and w its angular speed. So, if we use again conservation of energy, the final equation for h becomes:
$h= \frac{1}{2} \frac{v^2}{g} +2R + \frac{1}{2} \frac{I \omega^2}{mg}$
and as we can see, there is an additional positive term compared to the height we find for the sliding object, so the height at which the rolling object should be released is greater.