Angular Acceleration of a rod

TG3

66
0
1. The problem statement, all variables and given/known data
A uniform rod with of weight of 10 kg and length of 5.8 m is pivoted at its center and a small weight of mass 5.15 kg is rigidly attached to one end. You may neglect the size of the weight and assume it is located right at the end of the rod. The system is released from rest at a 37° angle. There are no external forces. What is the angular acceleration just after it is released?

2. Relevant equations

Torque = Angular Acceleration times Inertia
Torque = Force times Distance from pivot
Inertia of a uniformly weighted bar = 1/12 M L^2
I = Icm + M D^2

3. The attempt at a solution
Since the bar is pivoting about it's center, the net torque of the bar itself is zero. (One side balances the other out.)
The torque of the weight is 2.9 x 50.52 = 146.51
146.51 = I times angular acceleration.
The inertia of the bar = 1/12 (10) 5.8^2 = 28.033
146.5 / 28.033 = 5.226
This is wrong. Clearly I need to calculate the Inertia for the Weight, not just the bar, but since the problem says to neglect the size of the weight, I assume you have to treat it as part of the bar.
How do you calculate the inertia for a non-uniform bar?
 

Doc Al

Mentor
44,642
966
The torque of the weight is 2.9 x 50.52 = 146.51
That would be true if the bar were horizontal, but it's not. Consider the angle.
 

TG3

66
0
Ah yes. That did it. Thanks!
I'm also having trouble with a follow up problem:

What is the angular velocity when the rod is vertical?

The distance the rod falls is 1.15m, since sin 37 x 2.9 = 1.745 and 2.9- 1.74 = 1.15.
Final Velocity of an object falling in free fall is the square root of 2gy, so 9.81 x 1.15 x 2 = 4.7598
Angular Velocity = Tangential Velocity / radius
W = 4.7598 / 2.9
W = 1.64, according to my (wrong) calculations.
 

Doc Al

Mentor
44,642
966
The distance the rod falls is 1.15m, since sin 37 x 2.9 = 1.745 and 2.9- 1.74 = 1.15.
OK. The end of the rod falls that distance.
Final Velocity of an object falling in free fall is the square root of 2gy, so 9.81 x 1.15 x 2 = 4.7598
Nothing is in free fall here. (The rod rotates about a pivot, which exerts a force on it.)

Instead, consider conservation of energy. Find the rotational KE of the system.
 

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