HELP! Angular Acceleration

  • Thread starter justagirl
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  • #1
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A dumbbell consists of a slender rod of negligible mass and small spheres attached to each end with mass 12-kg and 6-kg respectively. It is pivoted at its center about a fixed horizontal frictionless axle and initially held in place horizontally. The length of the rod is 50 centimeters and connects the centers of the two spheres. The dumbell is then released.

What is the magnitude of angular acceleration immediately upon release?

I wrote down torque = I(alpha)
alpha = torque / I = 12(0.5) - 6(0.5) / I

I am stuck here. And what would "I" be in this case since the rod is neglible mass and we dont know the radius of the spheres?

Thanks!
 

Answers and Replies

  • #2
Doc Al
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justagirl said:
I am stuck here. And what would "I" be in this case since the rod is neglible mass and we dont know the radius of the spheres?
Treat the spheres as point masses. What's the rotational inertia of a point mass fixed a certain distance from an axis?
 
  • #3
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Okay... treating them as point masses:

alpha = torque / I

= (m1)gr - (m2)gr / m1(r^2) + m2(r^2)

= 12(9.8)(0.25) - 6(9.8)(0.25) / ((.25^2) (m1 + m2))

= 13.1

Is that right? So the position of the center of mass doesn't really matter in this case?

Thanks!
 
  • #4
Doc Al
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justagirl said:
= 13.1

Is that right?
Yes, in units of radians/sec^2.
So the position of the center of mass doesn't really matter in this case?
I'm not sure what you mean. If you had wished, you could have calculated the torque by considering the mass as concentrated at the center of mass.
 

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