What is the distance travelled by a rolling object which is released from an inclined plane?

Question

Suppose that a solid sphere with moment of inertia $I=\frac{2}{5}mr^2$ is rolling down an inclined plane. It's velocity at the end can be calculated using $mgh=\frac{1}{2}mv^2 + \frac{1}{2} I w^2$. Using $\frac{v}{r}=w$ we can get the final velocity as $v=\sqrt \frac{10gh}{7}$. If the sphere then proceeds to continue rolling on a surface with this as the initial velocity, it will eventually come to rest due to friction. Suppose that the coefficient of friction is $\mu$, what will be the distance travelled by the sphere when it comes to rest?

Answer 1

If the only friction at play is static friction, then it won't come to rest. It will just continue rolling with constant $v$ and $\omega$.

Answer 2

For pure rolling (no slipping) of a rigid body, friction does no work and the body will not slow down. With slipping the rigid body will slow down. No body is perfectly rigid and for rolling over a long distance the body will slow down due to friction. Please see my response to A bowling ball on an infinitely long track on this exchange. Also, see the reference rolling resistance ^{[2], [3]} noted by @ABC to this earlier question. As this reference discusses, for rolling friction the frictional force can be modeled as $\mu_r N$ where $\mu_r$ is the low coefficient of rolling friction and $N$ is the normal force. The distance traveled by the center of mass of the body is that of a body with an initial velocity (at the bottom of the incline) acted upon by a constant force, and can be calculated using standard equations for motion given a constant force.

Answer 3

The mechanical work done by the friction will be equal with the initial kinetic energy.

$\Delta E = \sum W$

The initial kinetic energy of the rolling sphere comes from the linear motion and the rotating motion.

$E = \frac{1}{2}mv^2 + \frac{1}{2} I\omega^2$

The mechanical work of the rolling friction is the power of the frictional force and the distance ($s$).

$W = Fs$

The frictional force is the power of the rolling resistance coefficient ($C_{rr}$) and the normal force.

In our case $\mu$ has been given which normally the sign of the slip friction coefficient. Assuming that $\mu$ is in our case the coefficient of the rolling resistance we can write.

$F = N\mu$

where $N$ is the weight force of the sphere ($mg$).

Putting it all together:

$\frac{1}{2}mv^2 + \frac{1}{2} I \omega^2 = mgs\mu$

from which you can express $s$.

In case of $\mu$ is NOT the rolling resistance coefficient, please, see John's answer about $C_{rr}$.

Answer 4

In your calculation of the kinetic energy of the sphere rolling down the incline you assumed no friction related energy losses. That means you have (1) pure rolling without slipping for no kinetic friction losses and (2) a perfectly rigid sphere and incline for no rolling resistance losses. Static friction does no work, it just enables rolling without slipping down the incline. so It involves no energy losses.

So unless the horizontal surface is not rigid so that there is now rolling resistance, the sphere could theoretically roll forever in the absence of air resistance..

The problem is, there is no such thing as perfectly rigid bodies. Rolling resistance will bring it to a stop, as well as reduce the kinetic energy acquired rolling down the incline.

Hope this helps.