A flywheel question...?

Question

A flywheel of mass 40kg and radius 2m, positioned so that it rotates about an horizontal axis...?

is initially at rest. A length of light string is wrapped around the flywheel with a mass of 2kg attached to the free end. The mass is initially 4m above the ground. As the mass drops, the string unwinds and the flywheel accelerates.

The question states:

At the instant the attached mass reaches the ground, a pair of brake blocks are applied and the flywheel is brought to rest in 0.2s. Given that the brake blocks are located 1.6m from the centre of gravity of the flywheel, and the coefficient of friction is 0.9, determine the normal reaction force between the flywheel and each brake block.

Usually if use of moment of inertia is required, it will quote the equation needed, however in this instance it hasn't. Is there anyway to do this without?

Thanks :)

Answer 1

You can model the flywheel as a disk, which has I=.5*m*r^2

I will use conservation of energy to answer the question.

The potential energy lost, which gets converted to kinetic energy of the falling mass plus the rotational energy of the spinning flywheel is
m*g*h
in this case
2*9.81*4
78.48 joules
At the instant the mass hits the ground
the falling mass has energy .5*m*v^2
and the spinning flywheel has energy
.5*I*w^2

since the string is at the edge of the flywheel, the v=w*r
or v^2=w^2*r^2

so the energy of the mass is
.5*2*w^2*4 or 4*w^2
and the energy of the wheel is
.5^2*20*4*w^2 or 20*w^2

back to balancing the energy
24*w^2=78.48
w=sqrt(78.48/24)
w=1.81 rad/sec

now that we have w, we can compute the reaction force of the brakes
since T=I*alpha
and w(t)=w0-alpha*t
t=.2
w(.2)=0
so
alpha=1.81/.2
=9.05

T=.5*20*2^2*9.05
T=362

Now, to relate T to the reaction force,
R*µ*1.6=362
R*µ=226.25
Since there are two brake blocks, the R*µ for a single block is
226.25/2=113.125
since µ=.9, R=125.7 N

j