20. A solid sphere of mass M, and radius R, rolls down from the top of the incli

Question

20. A solid sphere of mass M, and radius R, rolls down from the top of the incline of length L. The incline makes an angle with horizontal ground·What will be the velocity of the centre of mass (C.M) of the sphere when it reaches at the bottom? (a) u = V (3/7)gLsírb) (b) u = V (10/7)glstra (c) u = V (10/7)(L/g)sind (d) none of them 21. A hallow spherical shell M-2 Kg, and radius R-0.01m, rolls down without slipping from the top of the incline of length L-100 m . The incline makes an angle 0-30 with horizontal ground .What will be the velocity of the centre of mass (C.M) of the sphere when it reaches at the bottom Given that: Moment of inertia of hollow spherical sphere about an axis passing through the center of mass is (2MR2/3 (a) 24.25 m/s (b) 55.85 m/s (c) 55.2 m/s (d)26.5 m/s (e)none of them

Explanation / Answer

Using law of conservation of energy

energy at the top of the incline = energy at the bottom of the incline

height of the top of the incline from the bottom is h = L*sin(theta)

energy at the top of the incline is m*g*h = m*g*L*sin(theta)

energy at the bottom of the incline is (0.5*m*v^2)+(0.5*I*w^2)

I is the moment of the inertia = (2/5)*M*R^2

w is the angular velocity = v/R

then

energy at the bottom of the incline is (0.5*M*v^2)+(0.5*(2/5)*M*R^2*(v/R)^2) = (0.5*M*v^2)+((2/10)*M*v^2) = (7/10)*M*V^2

then

M*g*L*sin(theta) = (7/10)*M*v^2

M cancels on both sides

v = sqrt((10/7)*g*L*sin(theta))

so the answer is b) v = sqrt((10/7)*g*L*sin(theta))

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