HSC Science (Electronics) 12th Board ExamMaharashtra State Board
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Derive an Expression for Kinetic Energy - HSC Science (Electronics) 12th Board Exam - Physics

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Question

Derive an expression for kinetic energy, when a rigid body is rolling on a horizontal surface without slipping.Hence tind kinetic energy for a solid sphere.

Solution

Total Rolling kinetic energy = Translational K. E. + Rotational K. E.

`=1/2 MV^2+1/2 Iomega^2`

But

`omega=V/R `

`Total Rolling K.E = 1/2 MV^2+1/2 I(V^2/R^2)`

`"for solid sphere" I=2/5 MR^2`

`Total Rolling K.E.=1/2 MV^2+1/2 (2MR^2)/5(V^2/R^2)`

`=1/2 MV^2+1/5 MV^2`

`=7/10 MV^2`

  Is there an error in this question or solution?

APPEARS IN

 2012-2013 (March) (with solutions)
Question 3.3 | 3.00 marks

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Solution Derive an Expression for Kinetic Energy Concept: K.E.(Kinetic Energy) and P.E.(Potential Energy) in S.H.M..
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