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

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).

$fcp = m \times acp \\ fcp = m \times {v}^{2} \div r \\ fcp = 2 \times {5}^{2} \div 0.75 \\ fcp = 2 \times 25 \div 0.75 \\ fcp = 50 \div 0.75$
$fcp = 66.666... = 66 \: newtons$

Answer: Last option, 66 N.

Answer 2

Answer:

the answer is 66 N

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).