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A rope of negligible mass is wound around a hollow cylinder of mass `3 kg` and radius `40 cm`. What is the angular acceleration of the cylinder, if the rope is pulled with a force of `30 N` ? What is the linear acceleration of the rope ? Assume that there is no slipping. |
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Answer» Here, M=3 kg, R=40 cm =0.4 m Moment of inertia of the hollow cylinder about its axis `l=MR^(2)=3(0.4)^(2)=0.48 kg m^(2)` Force applied F=30 N `therefore " Torque", tau=FxxR=300xx0.4=12 N-m`. If `alpha` is angular acceleration produced, then from `tau=I alpha` `alpha=(tau)/(I)=(12)/(0.48)=25 rad s^(-2)` Linear acceleration, `a=Ralpha=0.4xx25=10 ms^(-2)`. |
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