# Angular velocity of a block of mass

1. Apr 21, 2005

### WY

Hey I was doing a question and I need someone to check if my method is ok - or stop me if i'm on completely the wrong track!!!

A block of mass m is attached (with a massless string) to a wheel. Consider the bicycle wheel is not turning initially. The block is allowed to fall a distance of h. Assume that the wheel has a moment of inertia I about its rotation axis.

Find the angular speed of the wheel after the block has fallen a distance of h in terms of m,g,h, r(of the wheel) and I

I took the relative zero as where the mass is starting from:
so the wheel has KE = 0 and the mass has KE=0 and PE=0

Then after it has fallen a height of h:
Wheel: KE = 1/2(I*omega^2)
Mass: KE = 1/2mv^2 and PE=mgh

so i came up with the conservation of energy equation to be:
0 = 1/2(I*omega^2) + 1/2mv^2 + mgh
the v the weight will be travelling at will be the same angular velocity the wheel is turning i substituted omega in for v
the I rearranged it so that omega was the subject and got:
omega = sqrt((-2mgh)/(Im))

is this rite??? or ami completely wrong?? thanks in advance :)

2. Apr 21, 2005

### ramollari

$$\omega = \frac{v}{r}$$ ?

Also:

$$0 = \frac{1}{2}I\omega ^2 + \frac{1}{2}mv^2 - mgh$$

So don't use (+) but (-) for mgh.

3. Apr 22, 2005

### WY

hahah right thanks for that!