Question

A string is wrapped around a uniform cylinder of mass M and radius R. The cylinder is released from rest with the string vertical and its top end tied to a fixed bar. A uniform disk, cylindrical in shape, of mass M and radius R is suspended from a string. The top end of the string is attached to a fixed bar. The string extends down through a distance h before reaching the disk, whereupon the string winds around the cylindrical body of the disk. (a) Show that the tension in the string is one-third the weight of the cylinder. (Submit a file with a maximum size of 1 MB.) (b) Show that the magnitude of the acceleration of the center of gravity is 2g/3. (Submit a file with a maximum size of 1 MB.) (c) Show that the speed of the center of gravity is 4gh 3 1/2 after the cylinder has descended through distance h. Verify your answer to (c) with the energy approach. (Submit a file with a maximum size of 1 MB.)

Answer 1

Answer:

a.) W/3,  b.)2g/3   c.) (4gh/3)^0.5

Explanation:

First we have to find tension in terms of torque. To do that we have to find the moment of inertia of a rigid cylinder. From Wikipedia I get:

$I = 0.5mr^2$

We also know the equation for torque. Let T be the tension, r be the distance, I is the moment of inertia and alpha is angular acceleration (ignore theta because it is perpendicular)

$torque = I*\alpha = T*r$

We can then substitute a/r for α

Therefore we get:

$I*\frac{a}{r}=T*r$

Isolating T and substitute the moment of inertia in for I we get

$T = [0.5mr^2]a/r^2= 0.5ma$

There are two known forces acting on the cylinder, gravity and tension. The sum of these two forces gives us mass times acceleration (Newton's second law)

$F = ma => mg - 0.5ma = ma\\g=\frac{3}{2}a\\a = \frac{2}{3}g$

This allows us to plug acceleration back into Newton's Second Law:

$mg - T = m[\frac{2}{3}g]\\ T = \frac{1}{3} mg = \frac{1}{3}w$

w = the weight

For part b, we solved in part a:

$a = \frac{2}{3}g$

For part c, we use the conservation of energy. We know that the sum of energy in the system is zero.

$mgh + \frac{1}{2} mv^2 + \frac{1}{2}I(omega)^2 = 0\\(omega) = (\frac{v}{r})^2\\ I = \frac{1}{2} mr^2\\-gh +\frac{1}{2} v^2+\frac{1}{4} v^2 = 0\\gh = \frac{3}{4} v^2\\v = \sqrt{ \frac{4}{3}gh }$