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# Questions concern a classic figure-skating jump called the

## Problem 33MCQ Chapter 6

Physics: Principles with Applications | 6th Edition

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Physics: Principles with Applications | 6th Edition

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Problem 33MCQ

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 speed of the skater’s hand? A. 1 m/s B. 3 m/s C. 9 m/s D. 15 m/s

Step-by-Step Solution:

Solution Step 1 of 3 During leap the skater performs uniform circular motion.vIn order to calculate the speed of the skater’s hand or the linear velocity, first we need to calculate angular velocity of the hand. Angular Velocity() In a circular motion,the rate at which the body rotates or angular position of the body changes is called Angular velocity. 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. Relation between linear velocity(L) an d angular velocity() v = r…………...2 Where r is the radius of the rim, v is the linear velocity 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 Radius or the length of the arm, r= 0.8 m Since,for one complete revolution skater covers 2rad distance, Therefore for 1 and half revolution Displacement d=(1.5)2rad d=3 rad To find, Angular speed = Step 3 of 3 Using this data in equation 1, =(3 rad/0.5 s) =6 rad/s =18.84 rad/s

Step 3 of 3

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