Circular Motion/Work Homework: Finding Work

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The problem involves a spring with a force constant of 548 N/m attached to an 8 kg disk in uniform circular motion, stretching the spring by 4 cm. The key equations discussed include the centripetal force equation and the work-energy principle. The initial attempt to calculate work suggests that the centripetal force provided by the spring is responsible for the motion. However, it is concluded that the work done is zero because the centripetal force acts perpendicular to the direction of the disk's velocity. This indicates that no work is done on the disk in this scenario.
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Homework Statement


A spring has a force constant of 548 N/m and
an unstretched length of 6 cm. One end is
attached to a post that is free to rotate in the
center of a smooth table, as shown in the top
view below. The other end is attached to a
8 kg disk moving in uniform circular motion
on the table, which stretches the spring by
4 cm.
Note: Friction is negligible.

1. W = 8.26365 J
2. W = 4.13182 J
3. W = 1.3152 J
4. W = 0.00473472 J
5. W = 0 J



Homework Equations


Sum of forces = m*v^2/r
Work = ?


The Attempt at a Solution


I don't know.
 
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mmmm i think in this case the centripetal force is the one applied by the spring, so

(constant)x(4cm) = (m)x(v^2/r)

solve and get v, and then

work of the disk = (1/2)x(m)x(v^2) + (1/2)x(I)x(W^2)
= (1/2)x(m)x(v^2) + (1/2)x(1/2x m xR^2)x(W^2)
= (1/2)x(m)x(v^2) + (1/2)x(1/2x m x v^2)
= (3/4)x(m)x(v^2)
 
Last edited:
i was wrong, i think the work is 0 because the centripetal force is perpendicular to the direction of v
 
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