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 73.0 kg and is 7.70 m from the pivot. He is skating at a speed of 5.90 m/s. Determine the magnitude of the centripetal force that acts on him.
Centforce=mv^2/r
To determine the magnitude of the centripetal force acting on the skater, we can use the formula for centripetal force:
F = (m * v^2) / r
where F is the centripetal force, m is the mass of the skater, v is the velocity of the skater, and r is the distance between the skater and the pivot.
In this case, the skater's mass (m) is given as 73.0 kg, the velocity (v) is given as 5.90 m/s, and the distance to the pivot (r) is given as 7.70 m.
Substituting the given values into the formula, we have:
F = (73.0 kg * (5.90 m/s)^2) / 7.70 m
Now, we can solve the equation to find the magnitude of the centripetal force:
F = (73.0 kg * 34.81 m^2/s^2) / 7.70 m
F = 3199.93 kg⋅m/s^2 / m
F = 3199.93 N
Therefore, the magnitude of the centripetal force acting on the skater is approximately 3199.93 Newtons.