A uniform ball of radius r rolls without slipping down from the top of a sphere of radius…
Physics · JEE Advanced · NTA Exams — Rotational Motion
A uniform ball of radius r rolls without slipping down from the top of a sphere of radius R. The angular velocity of the ball when it breaks from the sphere is
- \(\sqrt{\frac{5 g(R+r)}{17 r^{2}}}\)
- \(\sqrt{\frac{\log (R+r)}{17 r^{2}}}\)
- \(\sqrt{\frac{5 g(R-r)}{10 r^{2}}}\)
- \(\sqrt{\frac{\log (R+r)}{7 r^{2}}}\)
Answer
(B) (R+r) 17 r^ 2
Sign up free on Edukali to view the step-by-step worked solution and practice thousands of similar questions.
Related practice questions
- A racing car is travelling along a track at a constant speed of 40 m/s. A T.V. camera men is recording the…
- A solid sphere of radius R has moment of inertia I about its geometrical axis. it is melted into a disc of…
- A metal rod of length L and mass m is pivoted at one end. A thin disk of mass M and radius R ( L ) is…
- A car is moving with a speed of 30 m/sec on a circular path of radius 500 m. Its is increasing at the rate of…
- A circular turn table of radius 0.5 m has a smooth groove as shown in fig. A ball of mass 90 g is placed…
- A spherical ball rolls on a table without slipping. The fraction of its total energy associated with rotation…
- The kinetic energy of a particle moving along a circle of radius R depends on the distance covered s as T =…
- A particle is moving along a circular path of radius 3 meter in such a way that the distance travelled…
More Rotational Motion questions · Browse all practice questions