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 69.0 kg and is 5.50 m from the pivot. He is skating at a speed of 6.10 m/s. Determine the magnitude of the centripetal force that acts on him.
To determine the magnitude of the centripetal force that acts on the skater farthest out, we can use the formula for centripetal force:
Fc = (m * v^2) / r
where:
Fc = centripetal force
m = mass of the skater
v = velocity of the skater
r = radius of the circular path
Let's plug in the values given:
m = 69.0 kg
v = 6.10 m/s
r = 5.50 m
Substituting these values into the formula, we get:
Fc = (69.0 kg * (6.10 m/s)^2) / 5.50 m
Simplifying, we have:
Fc = (69.0 kg * 37.21 m^2/s^2) / 5.50 m
Fc = 2571.49 N
Therefore, the magnitude of the centripetal force that acts on the skater farthest out is approximately 2571.49 N.
To determine the magnitude of the centripetal force that acts on the skater, we can use the centripetal force formula:
\(F_c = \frac{mv^2}{r}\)
Where:
- \(F_c\) is the centripetal force
- \(m\) is the mass of the skater
- \(v\) is the speed of the skater
- \(r\) is the distance between the skater and the pivot point
In this case:
- The mass of the skater, \(m\), is given as 69.0 kg
- The speed of the skater, \(v\), is given as 6.10 m/s
- The distance between the skater and the pivot point, \(r\), is given as 5.50 m
Now, we can substitute the values into the formula and calculate the centripetal force:
\(F_c = \frac{(69.0\, \text{kg})(6.10\, \text{m/s})^2}{5.50\, \text{m}}\)
Simplifying the equation:
\(F_c = \frac{(69.0 \times 37.21)}{5.50}\)
\(F_c = \frac{2568.49}{5.50}\)
\(F_c = 467.1\, \text{N}\)
Therefore, the magnitude of the centripetal force that acts on the skater in the "crack-the-whip" stunt is 467.1 N