Finding Torque for Angular Momentum Parking Gate

In summary, the student attempted to solve for torque using an equation from physics, but made an error in calculation that resulted in a wrong answer.
  • #1
Goofball Randy
21
0

Homework Statement


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Homework Equations



Torque = r X F

The Attempt at a Solution



r is 2.5m since that's the length of the red (minus 0.5), and F should be the weight, so ma? That gives something like (5 * 2.5/3) * 9.8 * 2.5, which doesn't give me an answer at all :(
 
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  • #2
You've included the weight of the short end, but as though it is at the mass centre of the long end.
 
  • #3
haruspex said:
You've included the weight of the short end, but as though it is at the mass centre of the long end.

Do I need to calculate them individually? The correct answer here is 49, and I think my approach as a whole is wrong.
 
  • #4
Goofball Randy said:
Do I need to calculate them individually? The correct answer here is 49, and I think my approach as a whole is wrong.
Yes, your error is more fundamental, I realized just after posting. You seem to be calculating moment of inertia, which is not what is asked for.
 
  • #5
haruspex said:
Yes, your error is more fundamental, I realized just after posting. You seem to be calculating moment of inertia, which is not what is asked for.

Can you point out where? I thought I was calculating torque.
 
  • #6
Goofball Randy said:
Can you point out where? I thought I was calculating torque.
You squared the distance and divided by 3. That happens the formula for the moment of inertia of a bar, but it is not in the (correct) equation you quoted.
 
  • #7
haruspex said:
You squared the distance and divided by 3. That happens the formula for the moment of inertia of a bar, but it is not in the (correct) equation you quoted.

I'm sorry, I don't understand. Are you saying I SHOULD be using inertia of a bar, I = 1/3 M r^2 ? Because right now I am literally just doing f = ma.
And if I need to use inertia, should I use the torque = Inertia * angular acceleration equation?
 
  • #8
Goofball Randy said:
I'm sorry, I don't understand. Are you saying I SHOULD be using inertia of a bar, I = 1/3 M r^2 ? Because right now I am literally just doing f = ma.
And if I need to use inertia, should I use the torque = Inertia * angular acceleration equation?
No, I said the equation you quoted is correct:
Goofball Randy said:

Homework Equations



Torque = r X F
But that's not what you did here:
Goofball Randy said:
like (5 * 2.5/3) * 9.8 * 2.5, which doesn't give me an answer at all :(
Yiu have two factors of 2.5 and a divisor of 3. That looks like a moment of inertia calculation to me.
 
  • #9
haruspex said:
No, I said the equation you quoted is correct:

But that's not what you did here:

Yiu have two factors of 2.5 and a divisor of 3. That looks like a moment of inertia calculation to me.

Oh!

It wasn't, that was a straight up r x F calculation. See, I was confused by the part where the bar is attached "0.5m from the end". So I figured that the actual mass would not be 5kg, but rather 5 * (2.5/3), since that would be the fraction of the bar that I'm actually looking at.
 
  • #10
Goofball Randy said:
Oh!

It wasn't, that was a straight up r x F calculation. See, I was confused by the part where the bar is attached "0.5m from the end". So I figured that the actual mass would not be 5kg, but rather 5 * (2.5/3), since that would be the fraction of the bar that I'm actually looking at.
Ok. In that case, your error is that you have treated the short end as simply not being there. It will produce a counter torque.
The easiest way is just to treat all the mass as being at the mass centre (which is valid for finding the effective weight).
 
  • #11
haruspex said:
Ok. In that case, your error is that you have treated the short end as simply not being there. It will produce a counter torque.
The easiest way is just to treat all the mass as being at the mass centre (which is valid for finding the effective weight).

I see, how clever!

The mass center would be 1m away from the pole-thingy.

So torque = 1 * 5 * 9.8

= 49.

Thanks so much!
 

1. What is angular momentum parking gate?

Angular momentum parking gate is a type of parking gate system that uses the principle of angular momentum to control the movement of vehicles in and out of a parking lot.

2. How does angular momentum parking gate work?

The gate is designed with a rotating arm that is connected to a central axis. When a vehicle approaches the gate, the arm rotates to allow the vehicle to enter or exit the parking lot. The momentum created by the rotating arm helps to control the speed and direction of the vehicle.

3. What are the benefits of using angular momentum parking gate?

Angular momentum parking gate offers several benefits, including improved efficiency, reduced waiting time, and increased safety. The use of angular momentum helps to control the speed and direction of the vehicle, making the process of entering and exiting the parking lot smoother and faster.

4. Is angular momentum parking gate suitable for all types of parking lots?

Angular momentum parking gate can be customized to fit different types of parking lots, including those with narrow entrances or limited space. Its flexible design allows for easy installation and integration with existing parking systems.

5. Are there any drawbacks to using angular momentum parking gate?

One potential drawback of angular momentum parking gate is its reliance on electricity and mechanical parts, which can lead to malfunctions or breakdowns. However, regular maintenance and proper installation can help to minimize these issues.

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