A customer service person in a department store is wrapping packages for custome

Question

A customer service person in a department store is wrapping packages for customers. The ribbon spool she uses has 13.7 m of ribbon wound on it. The radius of the spool is 13.5 cm and its moment of inertia is 0.35 kg · m2. As the spool turns, friction forces cause a torque of 0.15 N · m to act on it. The spool is initially at rest. The service person pulls on the spool with a constant tension of 13.0 N. Determine the time it takes to unwind 5.0 m of the ribbon, assuming there is no slippage and the mass of the ribbon is negligible.

Explanation / Answer

Net Torque = I = (0.35

Hand Torque = 13(0.135) = 1.755 N-m

Resistance to Hand Torque (friction torque) = 0.150 N-m

Net Torque on spool = 1.755 - 0.150 = 1.605 N-m

1.605 = (0.35)

= 4.5758 rad/s²

a = R =4.5758(0.135) = 0.6191 m/s²

d = 1/2at²

5.0 = (0.5)(0.6191 )t²

t² = 16.15

t = 4.019 sec= ANS

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