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A uniform solid sphere rolls on a horizontal surface at 20 ms–1. It then rolls up an incline having an angle of inclination at 30º with the horizontal If the friction losses are negligible, the value of height h above the ground where the ball stops is :

Piyush Peeyush , 9 Years ago
Grade 12th pass
anser 1 Answers
Shaswata Biswas
Let the mass of the sphere be m. Velocity, v = 20 m/s.
Let its radius, momen of inertia and angular velocity be r, I and \omega respectively.
Since the sphere is rolling 
Its KETotal = KETranlasional + KERotational
                      = \frac{1}{2}mv^{2} + \frac{1}{2}I \omega ^{2}
                      = \frac{1}{2}mv^{2} + \frac{1}{2}*\frac{2}{5}mr^{2} \omega ^{2}
                      = \frac{1}{2}mv^{2} + \frac{1}{5}mv ^{2}             [v = \omega r]
                      = \frac{7}{10}mv^{2}
Again, let it reaches a maximum height of h above the ground.
Then, KEbottom = PEtop
=> \frac{7}{10}mv^{2} = mgh
=> h = \frac{7v^{2}}{10g}
This is the maximum height reached.
THANKS
Last Activity: 9 Years ago
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