# Why isn't the answer to this Atwood Machine problem correct? [closed]

Given: $$m_1 = 2*m_2$$ $$m_{disc} = 3*m_1 = 6*m_2$$ $$h = 3R$$ (R is the radius of the cylinder at the top) Find $$v^2$$ in terms of $$g$$ and $$r$$, where $$v$$ is the velocity of the blocks when $$m_1$$ is just about to hit the floor. Assume that the string does not slip over the cylinder at the top, and that the cylinder does have a moment of inertia.

This was a problem that I found a while back, and I've been trying to solving it ever since. The correct answer is apparently $$\frac{4}{3}gr$$, but I keep on getting $$gr$$ as my answer. Here is my work:

We have the Conservation of Energy Equation $$m_1gh = \frac{1}{2}m_1{v_1}^2 + \frac{1}{2}m_2{v_2}^2+m_2gh+\frac{1}{2}I{\omega}^2$$

First, we substitute in the first given equation to get $$2m_2gh = m_2{v_1}^2 + \frac{1}{2}m_2{v_2}^2+m_2gh+\frac{1}{2}I{\omega}^2$$

Now, we calculate I to be $$I_{disk} = \frac{1}{2}mr^2 = 0.5(3*m_1)r^2 = 1.5m_1r^2$$.

Substituting back in and combining like terms: $$m_2gh = m_2{v_1}^2 + \frac{1}{2}m_2{v_2}^2+\frac{1}{2}(1.5m_1r^2){\omega}^2$$

Now, we realize that $$\omega = \frac{v}{r}$$, and also again use the first given equation: $$m_2gh = m_2{v_1}^2 + \frac{1}{2}m_2{v_2}^2+\frac{1}{2}(1.5(2m_2)r^2)(\frac{v^2}{r^2})$$

Now, since $$h = 3R$$, we get $$3m_2gr = m_2{v_1}^2 + \frac{1}{2}m_2{v_2}^2+1.5m_2v^2$$

Note that in this problem, v_1 = v_2, so $$3m_2gr = m_2{v}^2 + \frac{1}{2}m_2{v}^2+1.5m_2v^2$$ $$3m_2gr = 3m_2v^2$$

Thus, we get $$v^2 = \frac{3m_2gr}{3m_2}$$ So, $$v^2 = gr$$.

I've been stuck on this problem for a long time, and I don't know where my mistake is. Can somebody please point me in the right direction?

I believe your answer is correct. I assume there is a mistake in the textbook, and they meant to write that $$h=4R$$. Then the textbook answer would have been right.