# 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 } = 1/2mv^{2} + 1/2Iω^{2}**

**= 1/2v**[ω=v/r]

^{2}(m+ I/r^{2})### 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}**