Question

A rope of negligible mass is wrapped around a 225-kg solid cylinder of radius 0.400 m. The cylinder is suspended several meters off the ground with its axis oriented horizontally, and turns on that axis without friction. (a) If a 75.0-kg man takes hold of the free end of the rope and falls under the force of gravity, what is his acceleration?
m/s²

(b) What is the angular acceleration of the cylinder?
rad/s²

(c) If the mass of the rope were not neglected, what would happen to the angular acceleration of the cylinder as the man falls?

Answer 1

Solution-The torque produced by the tension of the rope is causing the cylinder to accelerate as it rotates.

Using the formula torque = T * r = T * 0.4
Torque = I * α

I = ½ * m * r^2 = ½ * 225 * 0.4^2 = 18
T * 0.4 = 18 * α
T = 45 * α
Now to find the man’s acceleration just replace α with a/r.

T = 45 * a/0.4 = 112.5 * a

Man’s weight. W = 75 * 9.8 = 735 N
Net downward force on the man = (his weight)- (tension)
Net force = 735 – T = 75 * a
T = 735 – 75 * a

solving the equations for a.
112.5 * a = 735 – 75 * a
187.5 * a = 735
a = 735 ÷ 187.5 = 3.92 m/s^2

Now just divide by r.
α = 3.92/0.4 = 9.8 rad/s^2

(c)The total mass of the system would be increased as the mass of the rope is included in the calculations.
Mass is the measure of the difficult of changing the velocity of an object.
As the mass of the rope increases, the acceleration of the system becomes more difficult.
Because the force that causing the acceleration is the weight of the man is
So the linear acceleration of man and rope, and the angular acceleration of the cylinder would decrease.