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Velocity of center of mass of spinning disk

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  • #1
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A solid uniform disk of mass 19.0 kg and radius 70.0 cm (.7 m) is at rest FLAT on a frictionless surface. A string is wrapped around the rim of the disk and a constant force of 35.0 N is applied to the string. The string does not slip on the rim. The string is being pulled. When the disk has moved a distance of 6 m, what is the speed of the center of mass?



Relevant equations (I think...):
I=1/2MR^2
FR=Iα
α=a/R


So far, I have:
FR=Iα
FR=1/2MR^2(a/R)
F=1/2Ma
a=2*F/M=3.33333 m/s^2

and then,
v^2=2*a*6m, since initial velocity is 0.
v=7.26 m/s

But this is wrong, and I'm not sure why. Please help!
 

Answers and Replies

  • #2
gneill
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20,792
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Newton's second law holds for translational motion, even if the net force results in a torque. The angular and linear components can be dealt with separately when they are not "coupled" via some friction or other mechanical mechanism (like a rolling object or a pulley and string).
 
  • #3
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Thank you. I'm not sure how I would use Newton's Second Law, due to the fact that the normal force and mg are perpendicular to the movement and therefore not contributing to the velocity.

I'm so frustrated :)
 
  • #4
gneill
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20,792
2,770
Thank you. I'm not sure how I would use Newton's Second Law, due to the fact that the normal force and mg are perpendicular to the movement and therefore not contributing to the velocity.

I'm so frustrated :)
mg and the normal force are equal and opposite, so their contribution vanishes. You only have to deal with the force applied via the string.
 
  • #5
3
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OH, thank you gneill!!!

Answer is:

F=ma
35N=(21kg)a
a=1.667 m/s^2

v^2=0 + 2(1.667)(7.9), 0 being the initial velocity
v=5.1 m/s
 

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