Question

What is the linear speed and centripetal acceleration of point X?

A horizontal disc rotates uniformly at a constant angular velocity about a central axis normal to the plane of the disc.

Point X is a distance 2L from the centre of the disc. Point Y is a distance L from the centre of the disc. Point Y has a linear speed v and a centripetal acceleration a.

Answer 1

Given,

`omega = "constant"`

Recall,

`a_"R" = nu^2/r`

This is the proportionality we want to use in this case,

`=> nu^2/a_"R" = r`, and by extension

`nu^2 propto r`, and `1/a_"R" = r`

`r_Y = L" " r_X = 2L`

What are the parameters if we increase the radius by a factor of `2`?

Hence,

`2r = (2nu)^2= 4nu^2`

the linear velocity of point `X` increases by a factor of `4`, and

`2r = 1/(1/2*a_"R")`

the centripetal acceleration of the point `X` is decreased by a factor of `2`.

I'm open to feedback if I made a mistake!