# Torque and angular acceleration

1. Nov 10, 2013

### fogvajarash

1. The problem statement, all variables and given/known data
A massless beam is supported only at one point, called the pivot point, as shown in the diagram. A block with mass m1 sits at the left end of the beam, a distance L1 from the pivot point. A block with mass m2 sits at the right end of the beam, a distance L2 from the pivot point (L2 > L1). Calculate all torques around the pivot point, remembering that positive is anti-clockwise.

Select Yes, No, Less than, Equal to or Cannot tell.

A. Given particular values of L1 and L2, does the angular acceleration depend only on m1/m2? (If it depends on the actual values of m1 and m2, put no).
B.Given particular values of m1, m2 and L1, is it always possible to choose L2 (with L2 > L1) such that the masses have no angular acceleration?)
C. If m1 = m2, will the masses have an angular acceleration?
D. If m1 L2 = m2 L1, is there a negative torque? (product of mass and distance)

2. Relevant equations

3. The attempt at a solution
I chose for the first one no, as it depends on the values for m1 and m2. The general equation i got for the angular acceleration was:

α = g(m1L1 - m2L2)/(m1L12+m2L22)

On the other hand, I think it's possible to choose an specific value for which the masses will have no angular acceleration (in a way that L2 = (m1L1)/m2). Then, if m1 = m2, the masses will have an angular acceleration as there exists a length difference, so i chose yes for this option. Finally, for the last one, i chose that there is a negative torque (net) as solving for the equation will give us a negative result, so yes.

Could someone please guide me in the right direction? Thank you very much.

2. Nov 10, 2013

### Simon Bridge

A: no
B: yes
C: yes
D: yes

Is that correct?
It is useful to clearly relate the answers to the question. Normally I have to tell people to show their reasoning.
... was there any place you found confusing or surprising?

In your reasoning for A you ended up with:
$$\alpha = \frac{m_1L_1-m_2L_2}{m_1L_1^2+m_2L_2^2}$$... can we express that in terms of the ratio of masses $m_r=m_1/m_2$

(The question is asking if you can get the same angular acceleration with different masses.)

Last edited: Nov 10, 2013
3. Nov 10, 2013

### fogvajarash

Yep, my answers were no, yes, yes, yes.

I was thinking that maybe for 2, the answer could be "no" as well, because we can choose L1 to be 0 and then we will always have an angular acceleration regardless of the value we choose for L2. And for the first one, now I'm not so sure about my answer. I've did what you said and i ended up with:

α = (mrL1 - L2)/(mrL21 + L22)

So i think that pretty much we can choose m1 and m2 that have the same ratio to get the same result (although wouldn't it be limited for just some cases?)

4. Nov 10, 2013

### Simon Bridge

As you say, L1 could be "given" as "L1=0" ... in which case L2=0=L1 would be the only balance point - violating the restriction that L2>L1.

Notice that it also says "given m1 and m2"? What if you are given m1=m2?
What if m2 > m1?

Perhaps express m1=m, m2=km (x is an arbitrary constant), L1=L, L2=Lx (x is the value to be found).
As long as x>0 then L2 > L1 ... so work out the equation for x(k) and find out if it is always positive.

Can you find a situation where that would not work?

5. Nov 11, 2013

### fogvajarash

Yeah now i realized my mistake. Then we would have yes, no, yes, yes?

6. Nov 11, 2013

### Simon Bridge

The last condition was that $m_2L_1=m_1L_2$
I thought I'd show you another approach:

$m_1L_1 < m_2L_2$ gives a negative (clockwise) torque.

The condition means that: $m_1 = m_2L_1/L_2$ sub that in and rearrange:

$m_2L_1^2/L_2 < m_2L_2\Leftrightarrow L_1 < L_2$ ... which is true.

Agrees with what you got. ;)
Well done.

Last edited: Nov 11, 2013