InterviewSolution
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InterviewSolution
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A solid cylinder rolls up an inclined plane of angle of inclination `30^(@)`. At the bottom of the inclined plane, the centre of mass of the cylinder has a speed of `5 m//s`. (a) How far will the cylinder go up the plane ? (B) How long will it take to return to the bottom ? |
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Answer» Here, `theta=30^(@), v=5 m//s` Let the cylinder go up the plane up to a height h. From `1//2 mv^(2)+1//2lW^(2)=mgh` `(1)/(2)mv^(2)+(1)/(2)((1)/(2)mr^(2))omega^(2)=mgh` `(3)/(4) mv^(2)=mgh` `h=(3v^(2))/(4g)=(3xx5^(2))/(4xx9.8)=1.913m` If s is the distance up the inclined plane, then as `sin theta =(h)/(s), s=(h)/(sin theta)=(1.913)/(sin 30^(@))3.856 m` Time taken to return to the bottom `t=sqrt((2s(1+(k^(2))/(r^(2))))/(g sin theta))=sqrt((2xx3.826(1+(1)/(2)))/(9.8 sin 30^(@)))=1.53s` |
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