Question

An athlete prepares to throw a 2.0-kilogram discus. His arm is 0.75 meters long. He spins around several times with the discus at the end of his out-stretched arm so that the discus reaches a velocity of 5.0 m/s. What is the centripetal force acting on the discus?

9.4 N
14 N
27 N
66 N

Answer 1

the answer is 66 N

Step-by-step explanation:

Centripetal Force (Fcp) = ?

His arm length = Radius (R) = 0.75 m

Discus velocity = Linear Velocity (V) = 5 m/s

Discus mass (m) = 2 kg

Centripetal Acceleration (Acp) = V^2/R or W^2 x R

In this case i will use the V^2/R formula, because it uses the discus velocity (V).

Answer 2

Centripetal Force (Fcp) = ?

His arm length = Radius (R) = 0.75 m

Discus velocity = Linear Velocity (V) = 5 m/s

Discus mass (m) = 2 kg

Centripetal Acceleration (Acp) = V^2/R or W^2 x R
In this case i will use the V^2/R formula, because it uses the discus velocity (V).


`fcp = m * acp \\ fcp = m * {v}^(2) / r \\ fcp = 2 * {5}^(2) / 0.75 \\ fcp = 2 * 25 / 0.75 \\ fcp = 50 / 0.75`

`fcp = 66.666... = 66 \: newtons`

Answer: Last option, 66 N.