A rope of negligible mass is wrapped around a 225-kg solid cylinder of radius 0.

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?

b) What is the angular acceleration of the cylinder?

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

Ok for a) I do not understand why it wouldn't just be the acceleration due to gravity, and b) my answer I got is off by more than 10% so idk what I am doing wrong

so someone PLEASE help!!! Explicitly describing each step would be much appreciated!

Explanation / Answer

a.
The moment of interia for the cyliner is
I = m.cr^2/2

ma = mg - I/r
ma = mg - m.cr^2/2 * a/r *1/r
ma = mg - m.c/2 a
(m + m.c/2)a = mg
a = mg/(m + m.c/2)
a = 3.924 m/s^2

b, = a/r = 9.81 rad/s^2 ....
Funny answer, I had to double check. It is because m/(m+m.c/2) = 0.400 --- just like the radius. Someone rigged the problem because they are unrelated.

c. It would increase because gravity would be pulling the rope down

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