# Comparison of movement of Disc and Hoop

## Main Question or Discussion Point

A disc and a hoop (ring) both of same mass and radius roll down an inclined plane of height "h". It is well known the disc has great velocity than that of the hoop at the bottom. I am fixed in the situation after they start rolling horizontally. Which will go farther? If the surface is frictionless, none will start and there is no question of stopping. But if they face friction, then I assume both will face same friction and same torque caused by this friction because they have same mass and radius. I can't quite figure out which will stop first and why? Please help.

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Delta2
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Gold Member
I don't think that friction alone can make either of the disk or the ring stop. IF we have pure rolling without slipping, then if you think about it the point of contact is always at zero velocity with respect to the ground so there is no kinetic friction. There needs to be some other external force that affects the c.o.m , and then the (static) friction can supply the necessary torque to reduce the angular velocity of the bodies.

To make it more interesting lets suppose that there is an external force equal to $mg\sin\theta$ that is parallel to the horizontal plane and opposite to the horizontal velocity they just gained from rolling down the inclined plane. $\theta$ is the angle of the inclined plane and m the mass of the disk or the ring. Also suppose the friction coefficient is the same in inclined plane and the horizontal plane. Which of the two bodies you think it will go further?

Last edited:
• Zahid Iftikhar
rcgldr
Homework Helper
Assuming zero losses while traveling the ramp, then both the disc and the hoop have the same total mechanical energy (linear + angular) = mgh (decrease in gravitational potential energy), just distributed differently. As pointed out already, if there are no opposing forces like drag or rolling resistance, then both will continue to role each at their own constant velocity. If there is an opposing force, if the force is speed sensitive, then the disc with its faster linear velocity experiences more opposing force than the hoop with it's slower linear velocity. If the opposing force is constant, then the energy reduction of both disc and hoop would be force times distance traveled, and since they both have the same initial total mechanical energy at the bottom of the ramp, they would both stop rolling at the same point. The disc would have a higher average velocity, so it would take less time to cover the same distance and stop compared to the hoop, but they would both end up stopped at the same point.

• Zahid Iftikhar and Delta2
Assuming zero losses while traveling the ramp, then both the disc and the hoop have the same total mechanical energy (linear + angular) = mgh (decrease in gravitational potential energy), just distributed differently. As pointed out already, if there are no opposing forces like drag or rolling resistance, then both will continue to role each at their own constant velocity. If there is an opposing force, if the force is speed sensitive, then the disc with its faster linear velocity experiences more opposing force than the hoop with it's slower linear velocity. If the opposing force is constant, then the energy reduction of both disc and hoop would be force times distance traveled, and since they both have the same initial total mechanical energy at the bottom of the ramp, they would both stop rolling at the same point. The disc would have a higher average velocity, so it would take less time to cover the same distance and stop compared to the hoop, but they would both end up stopped at the same point.
High Regards for the reply Sir.
When you say, if retarding force is speed sensitive, you mean, it is the air friction? otherwise the surface friction is constant for both the objects. So if we ignore air friction they will stop at the same time and cover equal distance, otherwise disc will stop first and cover less distance for F.d is constant?

jbriggs444
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2019 Award
So if we ignore air friction they will stop at the same time and cover equal distance
If we imagine the object rolling down one ramp, arriving at the bottom and then rolling up the opposite ramp then, in the absence of frictional losses, conservation of energy demands that it wind up at rest at the same height it started with.

The hoop will hit bottom with the same total kinetic energy as the disc, but it will be rolling more slowly at all times. Accordingly, it will take longer to complete the trip. It will wind up at the same place, but not at the same time.

• Zahid Iftikhar
rcgldr
Homework Helper
When you say, if retarding force is speed sensitive, you mean, it is the air friction? otherwise the surface friction is constant for both the objects. So if we ignore air friction they will stop at the same time and cover equal distance, otherwise disc will stop first and cover less distance for F.d is constant?
Surface friction doesn't slow down a rolling object. Rolling resistance slows down a rolling object,

https://en.wikipedia.org/wiki/Rolling_resistance

If you assume an ideal rolling resistance that results in the same opposing force on both wheels, they cover the same distance and stop, but the disc gets there sooner.

• Zahid Iftikhar