Question #274966

You whirl a 32 gram rubber stopper on a string in a horizontal circle above your head. The string has length 82 cm and completes 5 revolutions in 4.1seconds. What is the speed of the stopper in m/s? What is the tension on the string in newtons?


1
Expert's answer
2021-12-14T10:30:40-0500

T=t/n=4.1/5=0.82 (s)T=t/n=4.1/5=0.82\ (s)


ω=2π/T=23.14/0.82=7.66 (rad/s)\omega=2\pi/T=2\cdot3.14/0.82=7.66\ (rad/s)


v=ωr=7.660.82=6.28 (m/s)v=\omega r=7.66\cdot0.82=6.28\ (m/s) . Answer


F=maF=0.326.282/0.82=1.54 (N)F=ma\to F=0.32\cdot6.28^2/0.82=1.54\ (N) . Answer






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