Skating stunt "crack the whip"

Question

In a skating stunt known as "crack-the-whip", a number of skaters hold hands and form a straight line. They try to skate so that the line rotates about the skater at one end, who acts as the pivot. The skater farthest out has a mass of 83.8 kg and is 6.20 m from the pivot. He is skating at a speed of 6.75 m/s. Determine the magnitude of the centripetal force that acts on him.

Answer 1

Hi Hope!

Use your centripetal acceleration equation along with a force equation, with the radius being the length of the line (6.20m), the speed of the skater is 6.75 m/s, and the mass you need to use being the mass of that farthest out skater (83.8kg).

so, centripetal acceleration is

a_centripetal = v²/r

Multiply in by mass, and now you have the centripetal force because it's always true that force = mass X acceleration. Our Force Centripetal equation is:

F_centripetal = (mass)( a_centripetal ) = m (v²/r) a_centripetal

Putting in our values we get:

m (v²/r) a_centripetal = (83.8_kg)(6.75_m/s²)/(6.2_m)

Crank through those numbers--remembering that (6.75_m/s²) means multiply 6.75 by itself (not by 2) and you should get the correct answer in Newtons.

Let me know if it works out for you!

Cheers,

Laura

P.S. I got 615.8N. What did you get?