With a quarter-turn flick of the wrist, a student sets the frisbee rotating at 570 rpm . what is the magnitude of the torque, assumed constant, that the student applies?

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Ответ:

rodriguezjeslin086

23.12.2021

Physics

Momentum of the ball just after the release is 1.94 kg m/s

Explanation:

As we know that initially the spring and compressed against the ball

So here we will have

$\frac{1}{2}mv^2 = \frac{1}{2}kx^2$

now we have

$v = \sqrt{\frac{k}{m}} x$

Here we know that

mass = 0.5 kg

k = 120 N/m

x = 0.25 m

$v = \sqrt{\frac{120}{0.5}}(0.25)$

v = 3.87 m/s

now we can use formula of momentum here

P = mv

P = (0.5)(3.87)

P = 1.94 kg-m/s

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