Circular Motion/Work Homework: Finding Work

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SUMMARY

The discussion centers on calculating the work done by a spring with a force constant of 548 N/m attached to an 8 kg disk in uniform circular motion. The spring stretches by 4 cm, and the participant initially attempts to apply the centripetal force equation, concluding that the work done is zero due to the perpendicular nature of the centripetal force to the velocity vector. The participant evaluates various work values, ultimately recognizing that the work done by the spring does not contribute to the disk's kinetic energy in this scenario.

PREREQUISITES
  • Understanding of Hooke's Law and spring constants
  • Knowledge of centripetal force and uniform circular motion
  • Familiarity with work-energy principles in physics
  • Ability to manipulate equations involving mass, velocity, and radius
NEXT STEPS
  • Study Hooke's Law and its applications in circular motion
  • Learn about the relationship between centripetal force and work done
  • Explore the concept of potential energy stored in springs
  • Investigate the work-energy theorem in the context of rotational motion
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, circular motion, and energy concepts. This discussion is beneficial for anyone tackling problems involving springs and forces in rotational systems.

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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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