# Two problems involving rotational movement

1. Oct 31, 2007

### dzogi

1. First Problem

1. The problem statement, all variables and given/known data
A disc with mass of 50kg and radius of 20cm is rotating with a frequency of 480rpm, and after 50 seconds, as a result of the force of friction, it stops. What's the moment (momentum of force, torque) if during the rotation the disc made 200 rotations?

2. Relevant equations
$$M=I\epsilon$$

3. The attempt at a solution
$$R=0,2m; m=50kg; f=480min^{-1}=8Hz; t=50s; N=200;$$
$$w_0=\frac{2\pi}{1/8}=16\pi rad/s$$
$$w = 0rad/s$$
$$\epsilon=\frac{w-w_o}{t}=\frac{-16\pi}{50}rad/s$$
$$I=0.5mR^2=1$$

$$M=I\epsilon=-1.21924Nm$$

Is this correct? I can't see where to number of total rotations (angular distance) fits in, or maybe it's a distractor?

2. Second problem

1. The problem statement, all variables and given/known data
А rope is wrapped around a horizontal cylinder with $$M=17kg; R=0,1m$$. A bob with $$m=5kg$$ is attached at the end of the rope, at height of $$h=4m$$ above ground. The momentum of inertia of the cylinder is calculated with $$I=\frac{MR^2}{2}$$.

a) what's the speed of the bob when it hits ground?
b) calculate the total energy of the system.

2. Relevant equations

3. The attempt at a solution
I've solved b) pretty easily,
$$E=mgh=192,2J$$ which conforms to the solution in the book (so, the potential energy of the cylinder is ignored).

I've tried solving a) this way
$$mgh=\frac{mv^2}{2}+\frac{I\omega^2}{2}$$
If the liner velocity of the cylinder is equal to the speed of the bob at any given time, then we can substitute $$\omega=\frac{v}{R}$$
$$\vdots$$
Is this approach correct? I don't get the same solution with the one given in the book.

Last edited: Nov 1, 2007
2. Nov 1, 2007

### saket

Problem 1 seems fine.

Kinetic energy of the cylinder should be $$\frac{I\omega^2}{2}$$ and not $$\frac{M\omega^2}{2}$$, ain't it?

3. Nov 1, 2007

### dzogi

Yes, it was a mistype :)