QUESTION 27
A uniform disk of radius 0.40 m and mass 31.0 kg rolls on a plane without slipping with angular speed 3.0 rad/s. The rotational kinetic energy of the disk is __________. The moment of inertia of the disk is given by 0.5MR^{2}.
Given mass=31kg
Radius=0.40m
Angular velocity w=3rad/sec
Translational velocityv =wr=1.2m/s
Moment of inertia I =0.5MR²=2.48kgm²
Total kinetic energy of the disc will be equal to the sum of kinetic energy due to tranlational as well as due to its rotational motion
Kinetic energy due to translational motion=(1/2)mv²=(1/2)*2.48*(1.2*1.2)=22.32j
Kinetic energy due to roational motion=(1/2)Iw²
=(1/2)(1/2)(31)(0.4*0.4)(3*3)=11.16j
So total kinetic energy is 33.48joule
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