Rod and Angular Acceleration [SOLVED]

[rock.freak667]

[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 $\frac{\pi}{3}$

Homework Equations

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.

 

[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.

 

[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

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

 

[Doc Al]

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.)

 

[rock.freak667]

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]

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:
$$\tau = \vec{r}\times\vec{W} = rW\sin\theta$$

 

[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?

 

[Doc Al]

Looks good!

 

[rock.freak667]

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]

Let me look it over one more time.

 

[Doc Al]

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.

 

[rock.freak667]

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