0
$\begingroup$

Hi

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?

$\endgroup$
3
  • 2
    $\begingroup$ Book answers can be wrong. Show your teacher your work. $\endgroup$
    – Bill N
    Nov 16 '20 at 17:19
  • 1
    $\begingroup$ @BillN Are you willing to state that the OP's answer is correct? That would be useful to him when he talks to his teacher. $\endgroup$ Nov 16 '20 at 18:24
  • $\begingroup$ Yes. I didn't find a mistake. $\endgroup$
    – Bill N
    Nov 17 '20 at 0:50
3
$\begingroup$

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.

$\endgroup$
1
  • 4
    $\begingroup$ All too often, textbook publishers change the numbers provided in the exercises while forgetting to change the answers in the back of the book. $\endgroup$ Nov 16 '20 at 20:19

Not the answer you're looking for? Browse other questions tagged or ask your own question.