# Rod and angular acceleration

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[SOLVED] Rod...and angular acceleration

## Homework Statement

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 $\frac{\pi}{3}$

## 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.

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alphysicist
Homework Helper
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.

Homework Helper
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

A uniform rod AB, of length 2a and mass m, has a particle of mass $\frac{1}{2}m$ attached to B.

Last edited:
Doc Al
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.)

Homework Helper
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.)
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?

Doc Al
Mentor
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

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

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

Homework Helper
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?

Doc Al
Mentor
Looks good!

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

and I get

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

Doc Al
Mentor
Let me look it over one more time.

Doc Al
Mentor
I see the problem.
You can find the torque due to each weight separately and add them up:
$$\tau = \vec{r}\times\vec{W} = rW\sin\theta$$
In the above expression for torque, $\theta$ is the angle between $\vec{r}$ and $\vec{W}$, which is not $\pi/3$ but its complement.
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)=$\frac{mga\sqrt{3}}{2}$
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.

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