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lara31 [8.8K]
4 years ago
8

A student holds one end of a thread, which is wrapped around a cylindrical spool, as shown above. The student then drops the spo

ol from a height h above the floor, and the thread unwinds as it falls. The spool has a mass M and a radius R, and the thread has negligible mass. The spool can be approximated as a solid cylinder of moment of inertia I = 1 MR2. Express your answers in terms of M, R, h, and fundamental constants. At time t = 0, the spinning spool lands on the floor without bouncing and comes free from the thread. It continues to spin, but slips on the floor's surface while doing so. Assume a constant coefficient of sliding friction m. (c) Calculate the angular velocity of the spool as a function of time t.
Physics
1 answer:
lesya [120]4 years ago
7 0

by energy conservation we know that

KE or rotation + KE of translation = gravitational PE

now we have

\frac{1}{2}I\omega^2 + \frac{1}{2}mv^2 = mgH

also we know that

v = R\omega

now we have

\frac{1}{2}(\frac{1}{2}mR^2)\omega^2 + \frac{1}{2}m(R\omega)^2 = mgH

\frac{3}{4}mR^2\omega^2 = mgH

\omega = \sqrt{\frac{4gH}{3}}/R

now when it is rolling on ground the torque acting on it due to friction force is given by

\tau = R F_f

\tau = \mu mg R

\alpha = \frac{\mu mg R}{\frac{1}{2}mR^2}

\alpha = \frac{2 \mu g}{R}

now angular speed at any time is given as

\omega = \omega_i + \alpha t

\omega = \sqrt{\frac{4gH}{3}}/R -\frac{2 \mu g}{R} t

so above is the angular speed in terms of time "t"

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1 A 75-g ball is projected from a height of 1.6 m with a horizontal velocity of 2 m/s and bounces from a 400-g smooth plate supp
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Answer with explanation:

We are given that  

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Velocity after impact,v_1=\sqrt{2gh_2}=\sqrt{2\times 9.8\times 0.6}=3.4m/s

According to law of conservation of momentum  

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Substitute the values  

0.075\times 5.6+0=-0.075\times 3.4+0.4v_2

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Explanation:

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