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A mass of 10 kg connected at the end of a rod of negligible mass is rotating in a circle of radius 30 cm with an angular velocity of 10 rad/s. If this mass is brought to rest in 10 s by a brake, what is the magnitude of the torque applied ?A. 0.9 N-mB. 1.2 N-mC. 2.3 N-mD. 0.5 N-m |
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Answer» Correct Answer - A `omega_(1)=10 "rad s"^(-1), omega_(2)=0,t=10s` `:. alpha=(omega_(2)-omega_(1))/(t)=(0-10)/(10)=-1 "rad s"^(-2)` Negative sign means retardation Now, `I=MR^(2)=10xx(0.3)^(2)=0.9 "kg-m"^(2)` `:.` Torque, `tau=Ialpha=0.9xx(1)=0.9 "N-m"`. |
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