Rotational motion magnitude question -- Rod mounted on a pivot

[oddenforcer]

I've attempted this problem several times and keep getting 14 rads as the answer but its not an available answer choice... any help would be greatly appreciated.

The problem:

A uniform rod (mass = 2.0 kg, length = 0.6 m) is free to rotate about a frictionless pivot at one end. The rod is released from rest in the horizontal position. What is the magnitude of the angular acceleration of the rod at the instant it is 60 degrees below the horizontal?

Available answers: 15, 12, 18, 29, and 23 rads/s^2



Relevant equations:
The moment of inertia for this rod is (1/3)mL^2 according to wikipedia.

The Attempt at a Solution

Here's how I keep coming to 14 rads:
Calculated moment of inertia to be 0.24 using the above equation.
Calculated the angular acceleration using sin(60)*9.8 to get 8.49 m/s^2. Divided this by circumference of the circle the rod would make (8.49/3.77=2.25).
Multiplied 2.25 * 2∏ which gives me an answer of 14 rads/s^2.

As I typed this up I realized I made no use of the rods moment of inertia which I realize I probably need to use in order to solve the problem. I'm just not sure how to implement it.

Again any help would be greatly appreciated!

Thanks!

 

[Doc Al]

Calculated the angular acceleration using sin(60)*9.8 to get 8.49 m/s^2.

No. Instead, calculate the torque acting on the rod about the pivot.

 

[oddenforcer]

Thank you for the response.

I'll reattempt the problem with this information.

 

[oddenforcer]

I am still struggling. To calculate the torque don't I need the applied force? And to find the applied force don't I need the acceleration of the rod?

Sorry, physics is definitely a weak area of mine! Any more help would be appreciated!

 

[Nathanael]

EDIT:
Sorry, I thought the question said "find the angular velocity at 60 degrees" but now I realize it asks to find the angular acceleration.

Sorry about that.

 

[Doc Al]

To calculate the torque don't I need the applied force?

You need to know what forces act on the rod. Hint: There's just one force that you need consider.

And to find the applied force don't I need the acceleration of the rod?

Nope.

You'll use the torque and Newton's 2nd law (for rotation) to calculate the angular acceleration.

It's easier than you think.

 

[oddenforcer]

Still hasn't clicked yet.

I know the force is gravity. Would you calculate the force using acting on the rod as sin(60 degrees)*(moment of inertia)*9.8?

I did that... but I can't help but feel I am at a dead end. I'm not sure why I find this problem so confounding lol. It probably will seem easy once I figure it out... but until then I'm pulling my hair out!

 

[Nathanael]

I know the force is gravity. Would you calculate the force using acting on the rod as sin(60 degrees)*(moment of inertia)*9.8?

No, the force is simply m*g

How is torque usually calculated? What is the torque in this situation?

(Hint: if the rod is uniform, then the center of mass is halfway along the rod)

 

[oddenforcer]

So then the torque would be equal to sin(60)*9.8*2*0.3?

 

[Doc Al]

So then the torque would be equal to sin(60)*9.8*2*0.3?

Almost. The sin(60) is incorrect.

 

[Nathanael]

So then the torque would be equal to sin(60)*9.8*2*0.3?

Have you drawn a picture? You should always draw a picture when you're unsure.

60 degrees is the angle from the horizontal to the rod. Are you sure that's the correct angle to use?Edit:
Doc Al is faster

 

[DDSJ787878]

The previous comment is important. Always draw a picture for the problem. Doing that you know: Torque is (mg) times (L/2)* cos(60) . Note it is not sin(60). Also the moment of inertia is (1/3)M L^2. So Sum of the torques = I (alpha).