Inertia/ Rotational Energy?

Question

A horizontal disc of diameter 12.0 cm is spinning freely about a vertical axis through its centre at an angular speed of 72 rpm
A piece of putty of mass 5.0g drops onto and sticks to the disc at a distance of 4.0 cm from the centre. The angular speed reduces to 60 rpm. (a) Calculate the moment of inertia of the disc
No external torques are applied to the system during this process

(b) A constant tangential force is now applied to the rim of the
disc which brings the disc to rest in 6.0 s Calculate the
magnitude of this force

(c) Calculate the rotational energy of the system before and
after the putty is added to the disc.

Answer 1

The disk has I=.5*m*r^2

L=I*w

convert rpm to rad/sec multipl rpm by pi/30

72 rpm=7.54 rad/sec
60 rpm = 6.28 rad/sec

By conservation of momentum, and since the putty had no angular momentum prior to sticking to the disk,

L1=I*7.54

L2=I*6.28+.005*.04*6.28

L2=L1

I=(.005*.04*6.28)/(7.54-6.28)

I=0.001 kg m^2/sec

b) T=I*a a=T/I
T=F*.06 m

w=w0-a*t

0=6.28-(F*.06)/.001


6.27*.0001/.06=F

=0.1045 N

c) rotational energy is
.5*I*w^2

before

.5*0.001*7.54^2
=0.0284

after

.5*(0.001+.04*.005)*6.28^2

=0.0237

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