Rolling Motion of Ring, Disk, Sphere: tr<td<ts

[Avaron Cooper]

1. A ring , a disk and a sphere all of same mass and radius, with moments of inertia Ir, Id, Is respectively about their axes, roll down without slipping on an inclined plane from a given height. If the time taken for the ring, disk and sphere to reach the bottom of the plane are tr, td and ts respectively, then
1)tr<td<ts
2)tr=td=ts
3)tr>td>ts
4)tr>td=ts
5)tr>td<ts

Homework Equations

. torque=moment of inertia*angular acceleration

The Attempt at a Solution

I took the torque acting on the objects to be the same theefore moment of inertia to be indirectly proportional to angular acc.

Ir>Id>Is
therefore:
tr<td<ts

If anyone show where I went wrong, it'll be of great help.

 

[CWatters]

I took the torque acting on the objects to be the same theefore moment of inertia to be indirectly proportional to angular acc.

Can you explain? The moment of inertia of an object is normally a constant that depends on the objects shape (among other things). For example if you spin up a flywheel it's moment of inertia doesn't change.

 

[Avaron Cooper]

Can you explain? The moment of inertia of an object is normally a constant that depends on the objects shape (among other things). For example if you spin up a flywheel it's moment of inertia doesn't change.

Since the radii of the objects are equal and their masses are equal, i took the torque acting on them to be equal.

 

[Avaron Cooper]

Can you explain? The moment of inertia of an object is normally a constant that depends on the objects shape (among other things). For example if you spin up a flywheel it's moment of inertia doesn't change.

therefore considering torque= I * Angular acc. , I took angular acc is indirectly proportional to I

 

[paisiello2]

I took the torque acting on the objects to be the same theefore moment of inertia to be indirectly proportional to angular acc.

I think you meant to say "inversely proportional".

So if the accelerations are smaller for larger moments of inertia, what does that do to the time?

 

[Avaron Cooper]

I think you meant to say "inversely proportional".

So if the accelerations are smaller for larger moments of inertia, what does that do to the time?

oh.. Stupid me.. Smaller accelerations means longer time.. Thank you very much!

 

[CWatters]

 

[paisiello2]

 

[Avaron Cooper]

Thank you very much!