A solid sphere of mass 2 kg is making pure rolling on a horizontal surface with kinetic energy 2240 J. The velocity of centre of mass of the sphere will be ______ m s−1.
Question
A solid sphere of mass 2 kg is making pure rolling on a horizontal surface with kinetic energy 2240 J. The velocity of centre of mass of the sphere will be ______ m s−1.
Solution
In the case of pure rolling, the total kinetic energy (K.E) of the sphere is the sum of the translational kinetic energy and the rotational kinetic energy.
The translational kinetic energy (K.E) is given by (1/2)mv², where m is the mass and v is the velocity.
The rotational kinetic energy for a sphere is given by (1/2)Iω², where I is the moment of inertia and ω is the angular velocity. For a sphere, the moment of inertia I = (2/5)mr², and the angular velocity ω can be related to the linear velocity v by the relationship ω = v/r.
So, the rotational kinetic energy = (1/2)(2/5)mr²(v/r)² = (1/5)mv².
Therefore, the total kinetic energy = translational kinetic energy + rotational kinetic energy = (1/2)mv² + (1/5)mv² = (7/10)mv².
Given that the total kinetic energy is 2240 J and the mass m is 2 kg, we can solve for the velocity v:
2240 = (7/10)2v² 2240 = 1.4*v² v² = 2240 / 1.4 v² = 1600 v = sqrt(1600) v = 40 m/s
So, the velocity of the centre of mass of the sphere is 40 m/s.
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