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Q7.19     A hoop of radius 2m  weighs 100\: kg . It rolls along a horizontal floor so that its centre
             of mass has a speed of 20cm/s . How much work has to be done to stop it?

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Total energy of hoop is given by : 

                                             T\ =\ \frac{1}{2}mv^2\ +\ \frac{1}{2}Iw^2

Also the moment inertia of hoop is given by :    I\ =\ mr^2

We get,                              

                                             T\ =\ \frac{1}{2}mv^2\ +\ \frac{1}{2}(mr^2)w^2

And      v  = rw 

                                          T\ =\ \frac{1}{2}mv^2\ +\ \frac{1}{2}mv^2\ =\ mv^2

 

So the work required is   :     =\ mv^2\ =\ 100(0.2)^2\ =\ 4 J                                       

 

Posted by

Devendra Khairwa

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