# Rotational dynamics of platform

#### rayf01

I have no clue on this, im stumped anyone have an idea.

a platform is rotating at an angular speed of 2.2rad/s.A block is resing on this platform at a distance of 0.30m from the axis. The coefficient of static friction between the block and the platform is 0.75. Without any external torque acting on the system, the block is moved toward the axis. Ignore the moment of inertia of the platform and determine the smallest distance from the axis at which the block can be relocated and still remain in place as the platform rotates.

The answer is supposed to be 0.17m

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#### NateTG

Homework Helper
Because of the conservation of angular momentum, moving the block in increases the rotational speed of the system.

One way to solve this problem is to start by determining the angular speed as a function of the block's position.

With the angular speed you should be able to determine the centripetal acceleration of the block as a function of the block's position.

Since the force of friction is the only thing keeping the block on the platform, you can then calculate the maximum force of friction, and work your way back.

#### jamesrc

Gold Member
You need to find where the friction balances out the centripetal force. The most the friction force can be is &mu;N, where N is the normal force (equal to the weight in this problem) and &mu; is the coefficient of static friction. If you set this equal to the centripetal force at some distance r, you will find that the mass of the object cancels out and you can solve for r. Just remember that the speed of the platform, &omega;, is a function of r (the distance of the mass from the rotational axis). Use the conservation of angular momentum to set up a relationship between the speed of the platform and the radius of the block (the mass will cancel out here too). Hint: your final expression for r will have a cube root in it.

#### jamesrc

Gold Member
Yeah, what NateTG said. Looks like I was too slow on the draw.

#### rayf01

Thanks for the help.

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