Question

A dog sits 0.50 m from the center of a merry go round. if the dog's centripetal acceleration is 1.5m/s^2, how long does it take the dog to go around once?

Answer 1

The centripetal acceleration is equal to

`a_c = (v^2)/(r)`
where v is the tangential speed and r the radius of the orbit.

By using
`a_c = 1.5 m/s^2` and r=0.50 m, we can find the value of v:

`v= √(a_c r) = √((1.5 m/s^2)(0.5 m)) =0.87 m/s`

We want to find the time the dog needs to do one complete revolution. The length of one revolution corresponds to the perimeter of the orbit:

`L= 2 \pi r = 2 \pi (0.50 m)=3.14 m`

And so, the time needed is just the length of one revolution (the perimeter) divided by the velocity:

`t= (L)/(v)= (3.14 m)/(0.87 m/s)=3.60 s`