Question

Example 7.2 A Rotating Wheel Goal Apply the rotational kinematic equations. Problem A wheel rotates with a constant angular acceleration of 3.45 rad/s. Assume the angular speed of the wheel is 2.15 rad/s at t; = 0. (a) Through what angle does the wheel rotate between t = 0 and t = 2.00 s? Give your answer in radians and revolutions. (b) What is the angular speed of the wheel at t = 2.00 s? Strategy The angular acceleration is constant, so this problem just requires substituting given values into Equations 7.7 and 7.8. Solution (a) Find the angular displacement after 2.00 s, in both radians and revolutions. Use Equation 7.8, setting w; = 2.15 rad/s, a = 3.45 rad/s2, and t = 2.00 s. A0 = w;t +-at A0 = (2.15 rad/s)(2.00 s) + 3 (3.45 rad/s?) (2.00 s)? A0 = rad Convert radians to revolutions. 40 = (Aw rad)(1.00 rev/21 rad) A0 = rev (b) What is the angular speed of the wheel at t = 2.00 s? Substitute the same values into Equation 7.7. w = w;+ at = 2.15 rad/s + (3.45 rad/s) (2.00 s) rad/s w = Remarks The result of part (b) could also be obtained from Equation 7.9 and the results from part (a). Exercise 7.2 Hints: Getting Started | I'm Stuck (a) Find the angle through which the wheel rotates between t = 2.00 s and t = 3.50 s. rad (b) Find the angular speed when t = 3.50 s. rad/s

Answer 1

a go = loot + 2at? =(2.15) Gz) + 2 (3.45)( 2 ) - 11.2 radions and 11.2 (1.7825/ revolutions Co-60 talt = (2.15) +(3.45)(2) 19.05 feel see (6) 받르 ald Of=bott & te - (2.15) (3.5) + 2 (3.45) (3.5)2 - 28.65625 rad iso =-11.2 +28.65625 = 117.45625 rad اله وبو - يما (4) =(2.15) + (3.45) (3.5) = (14.225 fred/fe

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