A 7.50 kg disk with a diameter of 88.0 cm is spinning at a rate of 280 revolutions per minute. A block of wood is pressed into the edge of t

Question

A 7.50 kg disk with a diameter of 88.0 cm is spinning at a rate of 280 revolutions per minute. A block of wood is pressed into the edge of the disk with a force of 35.0 N. The coefficient of friction between the wood and the disk is 0.300. How long does it take the disk to stop spinning?

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RI SƠ 4 years 2021-07-22T22:11:26+00:00 1 Answers 17 views 0

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  1. diemthu
    0
    2021-07-22T22:12:48+00:00
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    Answer:

    Time taken by disk to stop spinning will be 4.60 sec

    Explanation:

    It is given mass m = 7.50 kg

    Diameter of the disk d = 88 cm = 0.88 m

    So radius of the disk r=\frac{d}{2}=\frac{0.88}{2}=0.44m

    Angular speed of the disk \omega =\frac{2\pi N}{60}=\frac{2\times 3.14\times 280}{60}=29.30rad/sec

    Force is given F = 35 N

    Coefficient of friction \mu =0.3

    Friction force will be equal to f=\mu F=0.3\times 35=10.5N

    So torque will be equal to \tau =Fr=10.5\times 0.44=4.62Nm

    Torque is also equal to \tau =I\alpha

    Moment of inertia of the disk I=\frac{1}{2}Mr^2=\frac{1}{2}\times 7.5\times 0.44^2=0.726kgm^2

    So 4.62=0.726\times \alpha

    \alpha =6.964rad/sec^2

    So time taken will be equal to t=\frac{\omega -0}{\alpha }=\frac{29.30-0}{6.964}=4.60sec

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