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Units And Measurement

Question
CBSEENPH11017677

How large is a torque needed to accelerate a wheel of moment of inertia, I = 2kgmfrom rest to 30rps in 20s? 

Solution

Here, we have
 straight ϖ subscript 0 equals 0
space
straight ϖ equals 30 space rps space equals space 60 space rad divided by straight s space

straight t equals 20 straight s comma space

straight I equals 2 space kgm squared
Now, using the equation of rotational motion, we have
                 straight ϖ equals straight ϖ subscript 0 plus straight alpha space straight t
therefore space space straight alpha space equals fraction numerator straight ϖ minus straight ϖ subscript 0 over denominator straight t end fraction space

or space space straight alpha space equals space fraction numerator 60 straight pi minus 0 over denominator 15 end fraction equals 4 straight pi space rad divided by straight s squared
Now comma space

space space straight tau equals space Iα space

space space space space equals 2 cross times 4 straight pi space

space space space space equals 8 straight pi space Nm