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?

Respuesta :

The centripetal acceleration is equal to
[tex]a_c = \frac{v^2}{r} [/tex]
where v is the tangential speed and r the radius of the orbit.

By using [tex]a_c = 1.5 m/s^2[/tex] and r=0.50 m, we can find the value of v:
[tex]v= \sqrt{a_c r} = \sqrt{(1.5 m/s^2)(0.5 m)} =0.87 m/s[/tex]

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:
[tex]L= 2 \pi r = 2 \pi (0.50 m)=3.14 m[/tex]

And so, the time needed is just the length of one revolution (the perimeter) divided by the velocity:
[tex]t= \frac{L}{v}= \frac{3.14 m}{0.87 m/s}=3.60 s [/tex]