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 80.0 kg and is 6.10 m from the pivot. He is skating at a speed of 6.80 m/s. Determine the magnitude of the centripetal force that acts on him.

Answer 1

Answer:

The magnitude of the centripetal force that acts on him

Explanation:

Given that,

Mass = 80.0 kg

Distance = 6.10 m

Speed = 6.80 m/s

We need to calculate the magnitude of the centripetal force that acts on him

Using formula of the centripetal force

$F_{c}=\dfrac{mv^2}{r}$

Where, F = force

m = mass

v = speed

r = distance

Put the value into the formula

$F_{c}=\dfrac{80.0\times(6.80)^2}{6.10}$

$F_{c}=606.4\ N$

Hence, The magnitude of the centripetal force that acts on him