# Initial Angular Acceleration

1. Jan 1, 2006

### chowyo123

A rod of length 2.4 m is pivoted at its center and masses of 2.3 kg and 3.1 kg are hung from its two ends. Find the initial angular acceleration of the rod when the system is released from a starting position of the rod that is horizontal.

The radius of the rod would be r=1/2(2.4).. which is 1.2.

Should I use the equation I=mr^2 or I=1/12ml^2? I=mr^2 is I=(2.3)(1.2)^2=3.312 and I=(3.1)(1.2)^2=4.464. I=1/12ml^2 is I=1/12(2.3)(2.4)^2=1.104 and I=1/12(3.1)(2.4)^2=1.488. Do i even need these for this problem?

Where do i go next?

2. Jan 1, 2006

### Staff: Mentor

Use Newton's 2nd law for rotational motion: $$\tau = I \alpha$$, where I is the total rotational inertia.

What's the mass of the rod?

3. Jan 1, 2006

### chowyo123

I=(2.3)(1.2)^2= 3.312
I=(3.1)(1.2)^2= 4.464

3.312+4.464= 7.776. = I..

is this right so far?

4. Jan 1, 2006

### Mindscrape

Is the rod assumed to be massless? It looks like you are on the right track, and now find the torque. I'd suggest, this time, doing all the algebra before you put in the numbers.

5. Jan 2, 2006

### chowyo123

The problem doesn't state the mass of the rod, just the masses on both ends.
How do i find the torque if i've only got the inertia?

6. Jan 2, 2006

### Staff: Mentor

The torque is caused by the weight of the hanging masses. Use the usual formula for finding torque.

7. Jan 2, 2006

### Mindscrape

Yeah, the usual formula of $$\tau_\textit{net} = FR_1 + FR_2$$ where the force would be the force of gravity and r the distance from the pivot to the perpindicular force (half the length of the rod).

Last edited: Jan 2, 2006
8. Jan 2, 2006

### chowyo123

so Tnet = (9.8 x 2.3)(1.2)+(9.8 x 3.1)(1.2)?

Tnet=F(m*a)R1+F(m*a)R2?

Last edited: Jan 2, 2006
9. Jan 2, 2006

### Staff: Mentor

One torque is clockwise; one counter-clockwise. They have different signs.

10. Jan 2, 2006

### chowyo123

so $$\tau_\textit{net} = FR_1 + FR_2$$ was not right?

11. Jan 2, 2006

### cscott

Yes, but you need to change the signs:

$$\tau_\textit{net} = F_1R_1 - F_2R_2$$

You have to subtract the torque caused by the lesser mass because it works in the opposite direction (like Doc said).

I added the subscripts to F because they're two different forces.

12. Jan 2, 2006

### chowyo123

So:
Tnet = (9.8 x 3.1)(1.2)-(9.8 x 2.3)(1.2)= 9.408? What's next?

$$\tau = I \alpha$$

$$\ 9.408=7.776\alpha$$ ??

Last edited: Jan 2, 2006
13. Jan 2, 2006

### Mindscrape

Well, yeah, it was plus a negative torque, which you could just write as minus.

Yes, that should give the right answer when you divide.

14. Jan 2, 2006

### chowyo123

15. Jan 2, 2006

### Mindscrape

Yeah, depending on what convention they wanted you to use for positive or negative torque then the sign will be different. But, it doesn't look like they ever told you what weight was on which end, so you are probably fine.

16. Jan 2, 2006

### chowyo123

ok, thanks for the help, I appreciate it much.