Two identical small balls, each having a mass of 1/2 kg , are attached to the ends of a light rod that is 50 cm long to form a dumb-bell. This dumb-bell rotates at an angular speed of 10 rad/s about a fixed axis passing perpendicular to the rod through its centre. For 0.10 s , an impulsive force with an average magnitude of 5.0 N is applied to one of the masses in the direction of its velocity. Determine the new angu

Step-by-step solution

The moment of inertia of the dumb-bell system about the axis of rotation is given by: I = 2mr^2 Substituting m = 0.5 kg and r = 0.5 2 = 0.25 m : I = 2 0.5 (0.25)^2 = 0.0625 kg m ^2 The initial angular momentum of the system is: L_i = I ₀ = 0.0625 10 = 0.625 kg m ^2/ s The torque produced by the impulsive force about the centre is: = F r = 5.0 0.25 = 1.25 N m The angular impulse applied to the system is: L = t = 1.25 0.10 = 0.125 kg m ^2/ s Since the force is applied in the direction of velocity, the angular momentu