Astring 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.)

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:

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)

We can then substitute a/r for α

Therefore we get:

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

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)

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

w = the weight

For part b, we solved in part a:

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

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j