# Homework Help: Rod and angular acceleration

1. May 9, 2008

### rock.freak667

[SOLVED] Rod...and angular acceleration

1. The problem statement, all variables and given/known data
A uniform rod AB, of length 2a and mass m, has a particle of mass 1
2m attached to B. The rod is
smoothly hinged at A to a fixed point and can rotate without resistance in a vertical plane. It is released
from rest with AB horizontal. Find, in terms of a and g, the angular acceleration of the rod when it
has rotated through an angle of $\frac{\pi}{3}$

2. Relevant equations

3. The attempt at a solution

My idea is that the torque due to the weight and the mass is the net torque and that should be the same as $I \alpha$ where I=moment of inertia of the entire thing.

I got $I=\frac{10ma^2}{3}$ but I don't know how to incorporate that angle it is rotated through.

2. May 10, 2008

### alphysicist

Hi rock.freak667,

How did you find the moment of inertia? It looks like you used that the moment of inertia of the particle is 2 m a^2, but I don't think that is correct. (Is the mass of the particle 2m? If so, there is an extra 1 in your post.)

To find the angular acceleration, draw a diagram when it's at that angle, and use the formula (net torque = I alpha) that you mentioned. From the diagram you can calculate the torques at that particular angle.

3. May 10, 2008

### rock.freak667

The MOI through the centre is given by $I_c=\frac{1}{12}ML^2$

so for the rod in question, the MOI would be (through the centre)

$$I_c=\frac{1}{12}(mg+\frac{mg}{2})(2a)^2=\frac{1}{2}mga^2$$

and then by the parallel axis theorem to get the MOI about the end.

$$I=I_c+Mr^2$$

$$I=\frac{1}{2}mga^2 + (mg+\frac{mg}{2})(a)^2=2mga^2$$

is that correct moment of inertia about the end?

And I typed part of the question wrong. It should read

Last edited: May 10, 2008
4. May 10, 2008

### Staff: Mentor

What's the moment of inertia of a rod about one end? Of the particle about that end? Add them.

(The moment of inertia depends on mass distribution not weight, so g should not appear.)

5. May 10, 2008

### rock.freak667

ok well then. Using the parallel axis theorem

$$I=\frac{1}{12}m(2a)^2+ma^2=\frac{4}{3}ma^2$$

and then adding the MOI of the mass(=$\frac{m}{2}(2a)^2=2ma^2$

then

$$I_{total}=\frac{4}{3}ma^2+2ma^2=\frac{10ma^2}{3}$$

So then when I draw the forces at that angle (weight of rod+weight of the mass at B) of $\frac{\pi}{3}$. How would I get the torque at that angle? Would it just be the total torque*sine of the angle?

6. May 10, 2008

### Staff: Mentor

Perfect.

You can find the torque due to each weight separately and add them up:
$$\tau = \vec{r}\times\vec{W} = rW\sin\theta$$

7. May 10, 2008

### rock.freak667

Torque due to Weight of rod=mgasin(pi/3)=$\frac{mga\sqrt{3}}{2}$
Torque due to weight of mass=(mg/2)(2a)sin(pi/3)=$\frac{mga\sqrt{3}}{2}$

Total torque = $mga\sqrt{3}$

so that

$$I_{total} \alpha=mga\sqrt{3} \Rightarrow \alpha=\frac{mga\sqrt{3}}{I_{total}}$$

are the torques correct?

8. May 10, 2008

### Staff: Mentor

Looks good!

9. May 13, 2008

### rock.freak667

Apparently not...I found that the answer is really $$\alpha= \frac{3g}{10a}$$

and I get

$$\alpha=\frac{9\sqrt{3}g}{20a}$$

10. May 13, 2008

### Staff: Mentor

Let me look it over one more time.

11. May 13, 2008

### Staff: Mentor

I see the problem.
In the above expression for torque, $\theta$ is the angle between $\vec{r}$ and $\vec{W}$, which is not $\pi/3$ but its complement.
I must have been asleep at the wheel and assumed you were taking the sine of the correct angle. Sorry about that!

Correct that and you'll get the expected answer.

12. May 13, 2008

### rock.freak667

ahhh..thanks..my diagram was wrong...I pit pi/3 between the rod and the hinge as the complement...thank you!