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| 1. |
A disc of mass 2 kg is rollng on a horizontal surface without slipping with a velocity of 0.1 m//s. What is it's rotational kinetic energy ? |
| Answer» Solution :Rotational K.E. =`1/2Iomega^2 =1/2(MR^2//2)Xv^2//R^2=1/4Mv^2=1/4xx2xx10^(-2)=5XX10^(-3)` J | |