Question

The solid cylinder and cylindrical shell in the figure have the same mass, same radius, and turn on frictionless, horizontal axles. (The cylindrical shell has lightweight spokes connecting the shell to the axle.) A rope is wrapped around each cylinder and tied to a block. The blocks have the same mass and are held the same height above the ground. Both blocks are released simultaneously. Which hits the ground first? Or is it a tie? Must explain why

The answer I got is

the left one pulley have the moment of inertia equal to I=mR^2/2 this is smaller than that of the right one. (I=mR^2).

so that lets use a simple conservation of energy equation.

Iw^2/2+mv^2/2=mg*h. at the same distance h.

because I_left < I_right so that w and v of left is larger than right. so that the left one will meet the ground sooner.

However, I do not understand why the cylinder on the right has more Inertia. Isn't

I = Sum of all Inertia points? Then wouldn't it mean that there is more inertia on the left?

By intuition, I did guess that the left did reach the ground faster but according to the conservation of energy equation, (and by my faulty logic), the inertia would be greater on the left and thus reach the ground slower?

Somebody please teach me.

Answer 1

As we know that

τ = r X F = Iα

Here α = Angular acceleration

So more the value of α , more will be the acceleration and thus more rapidly the velocity increases and hence more rapidly it strikes the Ground

So

Let M be the mass of the Block and m be the mass of the Cylinders and R is the radius of Cylinders

So

α = (MgR)/I

Here I = Moment of Inertia

As MgR is same for both

Therefore

Less is the Value of I , more will the value of α and Hence it will strike the ground First

As the Moment of Inertia, I of the Solid Cylinder is Less than that of Cylinderical Shell(Because in this mass distribution is farer ron the Axis of Rotation)

Therefore

So

Mass attached to the SOLID CYLINDER will strike the ground FIRST.