Question

How to solve this ??

A disc of radius 4m is rotating about its fixed centre with a constant angular velocity `omega = 2`rad/s ( in the horizontal plane). A block is also rotating with the disc without slipping. If coefficient of friction between the block and disc is 0.4, then the maximum distance at which the block can rotate without slipping is ( `g = 10m/s^2`)

Answer 1

`r_max =1"m"`

Explanation:

We have a disk of radius `R =4 "m"`.
The disk rotates at a constant angular velocity of `omega = 2 "rad"/"s"`.
The coefficient of friction between a block sitting on the disk and the disk is `mu = 0.4`.
Gravitational acceleration is taken to be `g = 10 "m"/"s"^2`.

We want to find the maximum distance away from the center of the disk that the block can sit at without slipping.

We need to consider the forces in the system. First, there is a centripetal force acting inwards toward the center of the disk. Second, there is a frictional force that is limiting the block's motion. The other two forces in the system are the block's weight on the disk and the normal force of the disk acting on the block. These two forces would be equal, since the block is sitting on the surface of the disk.

The centripetal force of the system is:

`=>color(blue)(F_c = m omega^2 r)`

The frictional force is:

`=>F_f = mu N`

where the normal force is equal to the weight of the block:

`=> N = W = mg`

Therefore the frictional force is:

`=>color(blue)(F_f = mu mg )`

The maximum distance `r_max` that we could place the block would be when:

`=>F_f < F_c`

So let's solve for where this threshold `r` value is by setting the forces equal.

`=>F_f = F_c`

`=>mu cancel m g = cancel m omega^2 r_max`

`=>mu g = omega^2 r_max`

`=> color(green)( r_max = (mu g)/(omega^2))`

Plugging in our values we get:

`=>r_max = ((0.4)(10))/(2)^2 = 1"m"`

So placing the block at a distance `1"m"` or greater from the center of the disk will cause it to slip.