# Blocks on a rotating disc connected by a spring

1. Jul 8, 2013

### Flibberti

To understand centripetal force due to friction better, I came up with this problem. I'm not entirely sure of my solution, though, so I'd be glad if someone else took it up too and suggested a way to work it out:

Two identical blocks, each of mass m, connected by a spring of spring constant k are placed on a disc of radius R rotating with an angular velocity ω. The natural length of the spring is l (<2R), and the spring is placed with its midpoint at the centre of the disc. The maximum possible frictional force on each block is inadequate to make the blocks move with uniform angular velocity ω at their positions. Describe the motion of the blocks- what will the final extension of the spring be? (The coefficient of friction between the blocks and the surface is n.)

Last edited: Jul 8, 2013
2. Jul 8, 2013

### Andrew Mason

Welcome to PF!

If the friction force between the blocks and the disk is 0, the blocks to not rotate and the spring will remain at length l. If the friction force is not 0, the blocks will eventually reach the speed of the the disk and the final extension of the spring will be maximum. You should be able to calculate that extension.

AM

3. Jul 8, 2013

### Alpharup

I think I have an equation;
Frictional force between the block and the disc--the spring force=the centripetal force on the block.Let l be the original length of the spring
The displacement can easily calculated by the equation
kx-μmg=m((l/2)+x)(ω^2)
Here x is the displacement, k the spring constant, μ the coefficient of friction between the mass and the disc.
This equation is applied only to one block.
The total elongation of spring is 2x..