two spheres of same mass m and radius r rolls down without slipping down an inclined plane . one of sphere is solid and the other is a thin  spherical shell . Explain which one will reach the bottom of the plane first , if released from the same point on the plane 

ANSWER – 

For a rolling down a spheres , total kinetic energy at the bottom of the plane

K=K_T + K_R  =  1/2mv² + 1/2Iω² = 1/2v²(m+ I/r²)          [ω=v/r]

If the velocities of solid and hollow spheres at any point on the plane be V_s and V_h respectively , we have for solid sphere

K_s=\frac{1}{2}V_s^{2}(m+\frac{2}{5}\frac{mr^{2}}{r^{2}})=\frac{7}{10}mV_s^{2}

And for the hollow sphere

K_h=\frac{1}{2}V_h^{2}(m+\frac{2}{3}\frac{mr^{2}}{r^{2}})=\frac{5}{6}mV_h^{2}

but at any point on the plane the total kinetic energy , when falling from the same point will be same for both the spheres 

K_s = K_h
\frac{7}{10}mV_{s}^{2}=\frac{5}{6}mV_{h}^{2}

\frac{V_s^{2}}{V_h^{2}}=\frac{50}{42}

V_S > V_h

Since solid spheres has greater velocity it will each the bottom first