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lara31 [8.8K]
2 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]2 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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Explanation:

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Suppose the rod is along the x-axis, let's look for the charge density per unit length, which is constant

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If we derive from the length we have

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Let us write the integral limits, the lower is the distance from the point to the nearest end of the rod "d" and the upper is this value plus the length of the rod "del" since with these limits we have all the chosen charge consider

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We take out the constant magnitudes and perform the integral

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Using   λ = Q/L

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let's use a bit of arithmetic to simplify the expression

     [ 1/d  - 1/ (d+L)]   = L /[d(d+L)]

The final result is

     E = k Q / [d(d+L)]

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Answer:

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