Question

4. A uniform solid sphere of radius 0.1 m rolls smoothly across a horizontal surface at a speed 0.5 m/s with kinetic energy 0.7 J. Find the mass of the sphere and its angular momentum

Answer 1

understanding the concept The kinetic energy of an object for pure rolling can be represented as k_rolling = k_r = 1/2 mv_cm^2 + 1/2 I_on omega^2 m = mass of sphere v_cm = velocity of eam of sphere I_cm = Moment of inertia of sphere about the centre of mass omega = angular velocity of sphere = V_cm/r r = radius of sphere I_cm for solid sphere = 2/5 mr^2 m = mass r = radius Now, given, r = 0.1 m V_cm = 0.5 m/s k_r = 0.75 = > omega = V_cm/r Assuming the case of pure rolling & the mass of sphere = m putting the values, k_r = 1/2 m v_cm^2 + 1/2 (2/5 mr^2) middot (V_cm/r)^2 = 1/2 (m V_cm^2 + 2/5 mr^2 middot V_cm^2/r^2) = 1/2 (mV_cm^2 + 2/5 m V_cm^2) k_r = 1/2 (7/5 middot m V_cm^2) = 7/10 middot V_cm^2 = > m = 10 k_r/7 V_cm^2 = 10 times 0.7/7 times (0.5)^2 = 4 kg