Questions concern a classic figure-skating jump called the axle. A skater starts the jump moving forward as shown in Figure, leaps into the air, and turns one-and-a-half revolutions before landing. The typical skater is in the air for about 0.5 s, and the skater’s hands are located about 0.8 m from the rotation axis. FIGURE 31 What is the approximate angular speed of the skater during the leap? A. 2 rad/s B. 6 rad/s C. 9 rad/s D. 20 rad/s

Solution Step 1 of 3 Angular displacement (d) When a particle moves along the circumference of the circle, the radius drawn from the center of the particle position rotates through a certain angle. The angle through which the radius turns is called angular displacement (d). The SI unit of angular displacement is radian (rad). Angular speed() The rate at which the body rotates or angular position of the body changes is called Angular speed. It is given by, d = Angular displacement / time taken= dt radians/Sec Where, is the angular speed,d is the angular displacement and dt is the time taken. Step 2 of 3 In the given problem, the skater performs uniform circular motion. He undergoes one and half revolution in 0.5 sec before landing. Given data, Time interval, dt= 0.5 sec Since,for one complete revolution skater covers 2rad distance, Therefore for 1 and half revolution Displacement d=(1.5)2rad d=3rad To find, Angular speed =