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# Answer: Questions concern a classic figure-skating jump ISBN: 9780130606204 3

## Solution for problem 32MCQ Chapter 6

Physics: Principles with Applications | 6th Edition

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Problem 32MCQ

Questions concern a classic figure-skating jump called the axel. 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 The skater’s arms are fully extended during the jump. What is the approximate centripetal acceleration of the skater’s hand? A. 10 m/s2 B. 30 m/s2 C. 300 m/s2 D. 450 m/s2

Step-by-Step Solution:

Solution Step 1 of During leap the skater performs uniform circular motion. 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……...1 Where, is the angular speed,d is the angular displacement and dt is the time taken. For a body in circular motion, the centripetal acceleration(a) is the rate of change of tangential velocity as in figure below Relation between centripetal acceleration(a) and angular velocity() 2 a = r………………..2 Where r is the radius or the arm length. 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=3rad 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

##### ISBN: 9780130606204

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