Question

A uniform sphere and a uniform cylinder with the same mass and radius roll at the same velocity side by side on a level surface without slipping. If the sphere and the cylinder approach an inclined plane and roll up it without slipping, will they be at the same height on the plane when they come to a stop? If not, what will be the percentage height difference?
Please, help

Answer 1

I got the ratio is 4/5 but it's wrong. The book solution is 7.1%

Answer 2

If you approached the problem like this, say that started at the top an let both objects go, one will reach the bottom before the other. The time difference will be because they had different accelerations. And they will have different accelerations because they have different moment of inertia. If we know the difference in acceleration, could we determine the difference in final velocity and height?

How did you approach?

Answer 3

I got ratio \(\dfrac{4}{5}=\dfrac{h1}{h2}\)

Answer 4

I don't count the time they move on horizontal surface, assume that they start at the bottom of the inclined plane

Answer 5

I got the same answer s the book, but it was a struggle.

The height up the ramp will be proportional to the kinetic energy of each object

The formula for KE is 0.5 m v^2 + 0.5 I w^2

where v is velocity, w is omega the angular and I is the moment inertia

I for spheres is 04. m r^2 and for cylinders 0.5 mr^2.

The KE difference is then 0.1 m r^2 w^2

and if you compare it against the sphere's KE rather than the cylinder's you get 7.1%

Lots of cancelling out once v expressed as r w.

Answer 6

typo I for spheres is 0.4 m r^2, value I used

Answer 7

thank you