4 initial angular spued of the skater is π1 = 3 rotations/rec. let πΌ1 be the initial rotational inertia of the skater. given that final rotational inertia of the skater is πΌ2 = πΌ1 2 let π2 bue the final angular spued of the skatur. From angular momentum conservation initial angular momentum = final angular momentum πΌ1 π1 πΌ1 π1 ⇒ π2 π2 π2 = πΌ2 π2 πΌ1 = π2 2 = 2π1 =2×3 = 6 rotations/rec. The linear speed is, π£ 2ππ π 2π(8) = 10 = 5.0265 m/s = 5.0 m/s( two significant figures ) = The angular speed is, π π£ π 2ππ = π π 2π = π 2π = 10 = 0.6283rad/s = 0.63rad/s( two significant figures ) = acceleration while moving around a circular track = v^2 / r it does not depend on the mass so the acceleration will be a) equal so option a) equal is the correct answer The torque is calculated as folloes: Torque = mg [r / 2 ] = [60 kg * 10 m/s2 * 3 m] /2 = 900 N.m Given that 1 (1) ⇒ π΅π¦ relation we know that. linear " velocity = radius (π) × angular (v) welocity (w) distance of particle. ο· For the particle located at centre of circle π£ centre = π × π = 0 × 2 ventre = om /s] ο· For the particle located at the rim of circle ππππ = π × π = 1 m × 2rad/s πrim = 2 m/s Thank You. Given Radius of circle π = 2 m Rate at which it us acrelerates. π£ = 10 m/s2 Angular arceleration of the object is given by πΌ= π π substituting the values = 10/2 = 5 repes radians /s2 The distance traveled by the object for a complete revolution is equal to the circemtenence of the cilcle. π = 2ππ 2π = 2ππ 2π π = = 0.318309886 m 2π π = 0.3183 m