A cylinder of radius 18.0 cm is free to rotate about a smooth horizontal axle?

Question

A mass of 0.380 kg hangs from a string wound round the cylinder.
(a) when the system is released from rest, the mass takes 2.0 s to fall through a height of 5 m. What is the moment of inertia of the cylinder?
(b) The cylinder is then replaced by a hollow cylinder of the same mass and dimensions. Discuss qualititively the effect on (i) the speed of the mass (ii) the rotational kinetic energy of the cylinder, when the mass falls through a height of 5 m

Let me break down the question one by one.

First of all, may I say that I alpha equals FR since Tau = I alpha
= F R?

Secondly, to get alpha, I need final angular velocity. wf

In order to get w, I shall get it from v = rw, v get from mgh = 1/2 m

v^2, therefore v = square root of 2gh, hence w = square root of 2gh

over r.

So the wi = 0 cos its from rest, t = 2. then alpha = 27.5 rad/s^2

The following shall be I = FR/alpha = mghr/27.5

Will I get I = 0.122, but the book answer says 0.0360.

(The question assumes we do not know that moment inertia for a
cylinder is 1/2
mr^2)

I need the answer for (a) in figures. Thanks

Answer 1

This is probably way late, but I’ll go for it anyway...
Your equation I = FR/alpha = mghr/27.5 is not quite right.
F is not mgh. And alpha is not 27.5, but on the right track.

Need to follow the potential energy turning into kinetic energy.
Potential energy lost by the mass dropping 5 meters is
mgh = (0.38kg)(9.8m/s2)(5m) = 18.62 Nm (or joules)
some of it is in the falling 0.38kg mass, (mv2)/2,
and some in the rotating cylinder, (I omega2)/2.

What’s omega at the 2 second mark?
Took 2 seconds to unwind 5m from a 0.18m radius,
that’s theta= 5 / 0.18 = 27.777 radians.
theta = (1/2) alpha t2, so alpha = 2theta/4s2 = 13.888 rad/s2.
omega = alpha t = (13.888rad/s2)(2s) = 27.777rad/s.
That also gives v = omega r = (27.777rad/s)(0.18m) = 5 m/s

So the KE in the falling mass is
(mv2)/2 = (0.38kg)(5m)2 / 2 = 4.75 Nm (or joules).
And our available mgh was 18.62 Nm, so we’re left with
18.62 - 4.75 = 13.87 Nm for the rotating cylinder KE,
which is (I omega2)/2 = I (27.777rad/s)2 / 2 = I (385.8) / s2

13.87 Nm = 13.87 kg m2/s2 = I (385.8) / s2 , or
I = (13.87 / 385.8) kg m2 = 0.360 kg m2

For part (b), the new cylinder "I" will be larger, so the mass will
accelerate more slowly, and so will the cylinder.
But you knew that already.

Answer 2

use acceleration a =radiusx alpha also a=2h/t(square)
where h is height and t is time of fall
do not use v= r w