Torque change of a rod with unequal masses

In summary, the conversation discusses the calculation of torque to bring two connected balls to a halt in 5.0 seconds. The method used involves calculating the center of mass and moment of inertia, but there was an error in using the wrong radius for one of the masses. After realizing the mistake, the correct angular velocity was used and the calculation was successful.
  • #1
Metalsonic75
29
0
A 1.0 kg ball and a 2.0 kg ball are connected by a 1.0-m-long rigid, massless rod. The rod is rotating cw about its center of mass at 20 rpm. What torque will bring the balls to a halt in 5.0 s?

I used (m1x1+m2x2)/(m1+m2) to solve for the center of mass (2/3 meters from the 1kg weight). Then I used that to solve for the moment of inertia, using I= m1r1^2+m2r2^2, and I got I=1. I know [tex]\alpha[/tex]=[tex]\tau[/tex]/I, and [tex]\alpha[/tex]=([tex]\omega[/tex]_final - [tex]\omega[/tex]_initial) / time. I plugged in my known values (0 rps, 1/3rps, and 5 seconds, respectively) and got -0.1333 for [tex]\alpha[/tex]. Then I multiplied [tex]\alpha[/tex] by I and got -0.1333, which is wrong. I don't know where I screwed up. Any help would be greatly appreciated. Thanks
 
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  • #2
In the calculation for the moment of inertia, you used the wrong radius for the wrong mass. The (2/3 m)^2 should be multiplied by 1.0 kg.
 
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  • #3
It doesn't seem to work... I took 1*(2/3)^2 + 2*(1/3)^2 and got (2/3). Then I multiplied -0.1333 by (2/3) (angular acceleration times I) and got 0.0888, which is still wrong.
 
  • #4
Metalsonic75 said:
It doesn't seem to work... I took 1*(2/3)^2 + 2*(1/3)^2 and got (2/3).
Metalsonic75 said:
Then I multiplied -0.1333 by (2/3) (angular acceleration times I) and got 0.0888, which is still wrong.
That is not the angular velocity, that is simply the frequencey, you need the angular frequency/velcoity.
 
  • #5
Oh crap...how did I miss that? I didn't understand what Hootenanny was saying until I realized that his / was just a slash, and not a division sign. Anyway to the OP, yes it's true...you're out by a factor of 2*pi as far as your angular velocity goes.
 
  • #6
Ah, yes. I multiplied 1/3rps by 2*pi and everything worked. Thank you!
 

1. What is torque change of a rod with unequal masses?

The torque change of a rod with unequal masses refers to the change in the rotational force exerted on the rod when there is a difference in the masses attached to each end of the rod.

2. How is torque change affected by unequal masses?

Torque change is directly affected by unequal masses. The greater the difference in masses between the two ends of the rod, the greater the torque change will be.

3. Can torque change be calculated for a rod with unequal masses?

Yes, torque change can be calculated using the formula T = F x r, where T represents torque, F represents force, and r represents the distance from the axis of rotation to the point where the force is applied. The difference in masses can be substituted for the force in this formula.

4. What factors can affect the torque change of a rod with unequal masses?

The main factor that affects torque change is the difference in masses. Additionally, the length of the rod and the distance between the masses can also impact the torque change.

5. How can torque change of a rod with unequal masses be applied in real life?

Torque change of a rod with unequal masses has practical applications in various fields such as engineering, physics, and mechanics. For example, it can be used to determine the stability and balance of structures or to calculate the force needed to rotate objects.

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