This experiment includes a rotating disk attached by a cable over a pulley to a

Question

This experiment includes a rotating disk attached by a cable over a pulley to a hanging mass (see Figure 2), which provides a torque on the disk. The mass hanger only moves vertically, and the disk just spins about its center, so there are only two relevant equations from Newton's second law. These two equations are connected through T, the tension in the string, and the relation between a and alpha. Because the string connects to the hanger and encircles the disk, the hanger's displacement is related to the disk's angular displacement by Delta x = r Delta theta. According to the definition of velocity and acceleration, we also have v = r omega and a = r alpha. Finally, rotating bodies have kinetic energy, which is of the same form as the kinetic energy of translational motion and is given by K_rot =1/2 I omega^2. The rotational kinetic energy is considered as a mechanical energy and should be incorporated into the energy conservation law. Considering the setup in Figure 2, please show that The string tension T = M_h(g - a), where is the mass of hanger. The disk's moment of inertia I = (g/a - 1) m_h r^2 (g/a - 1) > 0, so the moment of inertia above is always positive. The total kinetic energy of the system K_total = 1/2(I + m_hr^2) omega^2

Explanation / Answer

1.

considering the free body diagram of the hanger then

mg -T = ma

T = m(g-a)

2. angular acceleration of disk = a/r

tourque on disk = T * r

so T * r  = I *a/r

I = T * r^2 / a

substituting value of T from 1 we get the result .

3. gravity is the driving force here and tension the restraining force for the hanger so the acceleration will always be less than g .

4.

add the indiviual energy of the disk and the hanger

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