Force needed to stop rotating drum

[Eddie91]

Homework Statement

A rotating drum is stopped by a brake actuated by a hydraulic cylinder as shown. The moment of inertia for the drum is Io=20 kg-m2. The friction coefficient between the brake shoe and the drum is 0.40. The dimensions are as given in millimeters. Find the force (in N) applied by the hydraulic cylinder to stop the drum in 75 revs if the initial drum speed, omega1, is as given below.

Hint: Io is the moment of inertia for the drum. The initial rotational kinetic energy of the drum is 0.5Ioomega2. The rotation will stop when the work of the friction force (Work=Force x distance) is equal to this kinetic energy.

omega1[RPM] = 252;

http://www.mech.uq.edu.au/courses/engg1010/q/brake.jpg

Homework Equations

KE = 10*omega^2 (given)
Distance = Circumference * Revolutions
Ff = uN

and finally KE = Ff*d

The Attempt at a Solution

I can't seem to get this question right.

KE = 10*omega^2 (given)
distance = circumference * revolutions = 37.5pi
Ff = uN = 0.4*N

and if Ff*d = KE, then:
0.4 * N * 37.5pi = 10*omega^2
N = 0.21220659*omega^2

Then your mechanical advantage should end up being 3 if my maths is correct, so the answer should equal N/3.

However, I keep being told I'm way off, is there something I'm missing?

Cheers,

Eddie

 

[mighty2000]

Dear Eddie,

Thank you for your question. It seems like you are on the right track, but there are a few things that need to be clarified. First, the initial rotational kinetic energy of the drum is given by 0.5*I*omega^2, where I is the moment of inertia and omega is the angular velocity in radians per second. So, using the given values, we have:

KE = 0.5*20*(252*2*pi/60)^2 = 662.5 J

Next, the work done by the friction force is equal to the force times the distance it acts over. In this case, the distance is given by the circumference of the drum times the number of revolutions, so:

Work = F*d = F*(37.5*pi*75) = 279.6F

Setting these two equal to each other, we have:

662.5 = 279.6F

Solving for F, we get:

F = 2.367 N

Now, this is the force applied by the friction force on the drum. The force applied by the hydraulic cylinder will be equal and opposite to this, so the force applied by the cylinder will also be 2.367 N.

I hope this helps clarify things for you. Let me know if you have any further questions.