Find the angular acceleration of a hinged beam

[UselessLadder]

Hi,
I have the following problem that I'm lost on.

The answer is 41.4 rad/s² but I don't understand how to arrive at it.

Homework Equations

ƩT=I(alpha)
I=1/3 ML² for a rod with axis through one end.

The Attempt at a Solution

I solved the above equations to get (alpha)=T/I, but no matter what I try from here I get small numbers under 5.
I'm confused on a couple points: is the center of gravity in the middle of the rod or the tip? Should I add the mass of the box (since the cable is cut, not the rope)? And I'm really confused as to how to write the sum of torques.
I tried following this example, with and without the box, but it I'm getting small numbers not even close to 41. Could someone please explain clearly to go about solving this? Thank you!

 

[haruspex]

Pls post your working.

 

[UselessLadder]

I converted the masses to Newtons to get 833N for the beam and 4214N for the box. Then I used trig to get the length of the beam (3m) and the angles: 37degrees below the beam and 53 above. Then I tried this, based on an example from my class notes (assuming the center of the beam's gravity is in the middle):

,
substituting 5047N for m and 3 for L, then solving for alpha. But that's clearly wrong.
Then I tried following this example exactly as shown, with units as shown, but I got 4.4(without the box) and 3.9 (with the box's weight added). Not even close to 41.

 

[haruspex]

You're overlooking that the mass will accelerate too. Let the tension in the rope be T. Develop the free body equations for the beam and the mass separately. The relationship between the beam's angular acceleration and the mass's vertical acceleration will need a bit of care.
I recommend doing all working symbolically, only plugging in numbers as the final step. That goes for the angles too. It makes it much easier to follow the reasoning and to spot mistakes.

 

[Nickie]

Hello,

To find the angular acceleration of a hinged beam, we need to use the equation ƩT=I(alpha), where ƩT represents the sum of all the torques acting on the beam, I represents the moment of inertia, and alpha represents the angular acceleration.

First, we need to determine the moment of inertia of the beam. Since the beam is hinged at one end, we can use the equation I=1/3 ML^2, where M is the mass of the beam and L is the length of the beam.

Next, we need to determine the sum of all the torques acting on the beam. In this case, we have two forces acting on the beam: the weight of the beam and the tension in the cable. The weight of the beam can be represented by the force Mg, where g is the acceleration due to gravity. The tension in the cable can be represented by the force T.

Now, we can write the equation ƩT=I(alpha) as:
T - MgL/2 = (1/3)ML^2 * alpha

Since we are looking for the angular acceleration, we can rearrange this equation to solve for alpha:
alpha = (3T - 3MgL/2)/ML^2

Substituting the given values of T=100 N, M=10 kg, and L=2 m, we get:
alpha = (3*100 N - 3*10 kg*9.8 m/s^2*2 m/2)/(10 kg*2 m^2)
= (300 N - 294 N)/20 kgm^2
= 6 N/20 kgm^2
= 0.3 rad/s^2

This is not the answer we are looking for, as it is much smaller than 41.4 rad/s^2. This could be due to a few reasons:
1. The center of gravity of the beam is not in the middle. If the center of gravity is closer to the hinge, the moment of inertia will be smaller, resulting in a larger angular acceleration.
2. The mass of the box is not considered. If the mass of the box is added to the total mass of the beam, the moment of inertia will be larger, resulting in a smaller angular acceleration.
3. The direction of the forces is not taken into account. If the tension in the cable is acting