Help finding KE?

Question

A uniform disk of mass .11 kg and radius 21 cm is constrained to rotate on an axis about its center. Friction exerts a net torque of .0252 meter Newtons on the system when it is in motion. On the disk are mounted masses of 25 grams at a distance of 17.64 cm from the center, 10 grams data distance of 8.82 cm from the center and 50 grams at a distance of 5.04 cm from the center. A uniform force of .6 Newtons is applied at the rim of the disk in a direction tangent to the disk. The force is applied for 4 seconds with the disk initially at rest.

What is the KE?
At the system's final angular velocity what is the speed and KE of each of the masses mounted on the disk?

I need someone to walk me through this...not really looking for just an answer.

Answer 1

Torque=moment of inertia times angular acceleration. Much like translational physics where force equals mass times acceleration, the rotational frame of reference is analogous.

The first step is to compute the moment of inertia for the spinning object about the axis of rotation.

A disk rotating on it's center is
I=m*r^2/2
for the disk in this question
.11*.21^2/2
or 2.43/1000

The point masses add to the moment as follows
each point mass is m*r^2
.025**.1764^2
.01*.0882^2
.05*.0504^2
add these up and add to the disk to get
3.41/1000

This problem is set up to work the angular energy conservation where work is equal to torque times angular displacement

so, let's first find the angular displacement at time equals 4 seconds.
Again, analagous to translational frames of reference

th(t)=th0+w0*t+.5*al*t^2
where
th is the angle of displacement
th0 is the angle at t=0
w0 is the angular speed at t=0
al is the angular acceleration
since th0=0 and w0=0
th(4)=.5*al*16 or al*8

we compute al from above as
T=I*al
where T is torque

The net torque is
.6*0.21-0.0252 Nm
or
0.101 Nm

so al=1000*0.101/3.41
or 29.62
which makes th(4)=237 radians

So the work done by the torque is
237*.101 Joule
or 23.9 Joule
This is the kinetic energy at t=4 seconds.
I think this is what you meant by "final"
In reality, the friction torque will bring the disk to rest once the torque is removed.
I will continue to work at t=4

Let's check our work by finding w(4)
w(t)=w0+al*t
since w0=0 and al=29.62
w(4)=118 rad/sec

Similarly to translational physics, in a rotational reference frame
KE=.5*I*w^2
so
KE=.5*3.41*118^2/1000
=23.7 Joule
off slightly due to rounding
The good news is that the answer works both ways.


Compute the KE of each mass the same way. Since they are all mounted on the disk their angular speeds are all the same as the disk, only the individual moments change, which are all computed above.

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