1.

A hollow sphere of outer radius R is rolling down aninclined plane without slipping and attains a speed v_0 at the bottom. Now the inclined plane is made smooth and the sphere is allowed to slide without rolling. Now it attains a speed (5v_0)/4. What is the radius of gyration of sphere?

Answer»

`sqrt(2/5)R`
`3/4R`
`4/5R`
`sqrt(2/3)R`

Solution :Let the SLIDING acceleration be a. So rolling acceleration `=a/(1+K^2//R^2)`,
where K is the RADIUS of gyration.
Using `v^2-u^2=2as`
`RARR (5/4v_0)^2=2aS`….(i)
`v_0^2=(2aS)/(1+K^2//R^2)` …(ii)
From (i) and (ii) we get
`1+K^2/R^2=25/16` or `K^2/R^2=25/16-1`
`rArr K^2=(9R^2)/16` or `K=(3R)/4`


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