A massless string is wrapped around the equator of a solid sphere (mass M = 53.5

Question

A massless string is wrapped around the equator of a solid sphere (mass M = 53.5 kg, radius R = 0.263 m). Mary holds the free end of the string, and the sphere is released from rest, Assume:
- the sphere is always parallel to the floor
- the string is always perpendicular to the radius of the sphere
- the string does not slip over the sphere

a) Suppose first that Mary pulls the string upward at a speed that just prevents the center of mass of the sphere from falling as the string unwinds. Find:
- T, the magnitude of the tension in the string as the sphere turns: 1 N
- W, the net work that has been done on the sphere once it reaches angular speed ? = 27.2 rad/s: 2 J
- L, the length of the string that has been unwound when the sphere reaches an angular speed ? = 27.2 rad/s: 3 m

b) Suppose now that Mary holds the end of the string stationary, and the string unwinds as the sphere falls. Find:
- T, the magnitude of the tension in the string as the sphere turns: 4 N
- t, the time it takes for the falling sphere to reach an angular speed ? = 27.2 rad/s: 5 s A massless string is wrapped around the equator of a solid sphere (mass M = 53.5 kg, radius R = 0.263 m). Mary holds the free end of the string, and the sphere is released from rest, Assume:
- the sphere is always parallel to the floor
- the string is always perpendicular to the radius of the sphere
- the string does not slip over the sphere

a) Suppose first that Mary pulls the string upward at a speed that just prevents the center of mass of the sphere from falling as the string unwinds. Find:
- T, the magnitude of the tension in the string as the sphere turns: 1 N
- W, the net work that has been done on the sphere once it reaches angular speed ? = 27.2 rad/s: 2 J
- L, the length of the string that has been unwound when the sphere reaches an angular speed ? = 27.2 rad/s: 3 m

b) Suppose now that Mary holds the end of the string stationary, and the string unwinds as the sphere falls. Find:
- T, the magnitude of the tension in the string as the sphere turns: 4 N
- t, the time it takes for the falling sphere to reach an angular speed ? = 27.2 rad/s: 5 s

Explanation / Answer

v = ?r = 6.0346 m/s vf^2 = v0^2 + 2as s = 1.86 m

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