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Grade 12th passMechanics

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 :

Profile image of Piyush Peeyush
9 Years agoGrade 12th pass
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1 Answer

Profile image of Shaswata Biswas
9 Years ago
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