PROBLEM A wheel rotates with a constant angular acceleration of 3.50 rad/s. If the angular speed...
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PROBLEM A wheel rotates with a constant angular acceleration of 3.50 rad/s². If the angular speed of the wheel is 2.00 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 in revolutions. (b) What is the angular speed of the wheel at t = 2.00 s? (c) What angular displacement (in revolutions) results while the angular speed found in part (b) doubles? STRATEGY The angular acceleration is constant, so this problem just requires substituting given values into the proper equations. SOLUTION (A) Find the angular displacement after 2.00 s, in both radians and revolutions. Use the equation to the right, setting = 2.00 rad/s, a = 3.5 rad/s², and @i t = 2.00 s. Convert radians to revolutions. Substitute the same values into the equation to the right. Apply the time-independent rotational kinematics equation. ΔΘ = Substitute values, noting that Wf = 2W;. (B) What is the angular speed of the wheel at t = 2.00 s? Solve for the angular displacement and convert to revolutions. w₁t + 12/1at² = (2.00 rad/s)(2.00 s) + (3.50 rad/s²)(2.00 s)² = 11.0 rad A0 (11.0 rad) (1.00 rev/2л rad) = (C) What angular displacement (in revolutions) results during the time in which the angular speed found in part (b) doubles? w/2² -w₁² = 2a40 w = w;+ at = 2.00 rad/s + (3.50 rad/s²) (2.00 s) = 9.00 rad/s = 1.75 rev (2 x 9.00 rad/s)² – (9.00 rad/s)² = 2(3.50 rad/s²)40 A0 (34.7 rad)( = 1 rev 2л rad = 5.52 rev LEARN MORE REMARKS The result of part (b) could also be obtained using the equation ² = w;² + 2a40 and the results of part (a). QUESTION Suppose the radius of the wheel is doubled. Are the answers affected? If so, in what way? (Select all that apply.) The angular speed at t = 2.00 s is the same. The angle rotated through from t = 0 to t = 2.00 s is greater. The angular speed at t = 2.00 s is greater. The angular speed at t = 2.00 s is smaller. The angle rotated through from t = 0 to t = 2.00 s is the same. The angle rotated through from t = 0 to t = 2.00 s is smaller. PRACTICE IT Use the worked example above to help you solve this problem. A wheel rotates with a constant angular acceleration of 3.70 rad/s2. Assume the angular speed of the wheel is 2.25 rad/s at ti = 0. (a) Through what angle does the wheel rotate between t = 0 and t = 2.00 s? Give your answer in radians and revolutions. rad rev (b) What is the angular speed of the wheel at t = 2.00 s? rad/s (c) What angular displacement (in revolutions) results while the angular speed found in part (b) doubles? rev EXERCISE Use the values from PRACTICE IT to help you work this exercise. (a) Find the angle through which the wheel rotates between t = 2.00 s and t = 3.80 s. rad (b) Find the angular speed when t = 3.80 s. rad/s HINTS: GETTING STARTED I I'M STUCK! (c) What is the magnitude of the angular speed five revolutions following t = 3.80 s? rad/s Need Help? Read It PROBLEM A wheel rotates with a constant angular acceleration of 3.50 rad/s². If the angular speed of the wheel is 2.00 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 in revolutions. (b) What is the angular speed of the wheel at t = 2.00 s? (c) What angular displacement (in revolutions) results while the angular speed found in part (b) doubles? STRATEGY The angular acceleration is constant, so this problem just requires substituting given values into the proper equations. SOLUTION (A) Find the angular displacement after 2.00 s, in both radians and revolutions. Use the equation to the right, setting = 2.00 rad/s, a = 3.5 rad/s², and @i t = 2.00 s. Convert radians to revolutions. Substitute the same values into the equation to the right. Apply the time-independent rotational kinematics equation. ΔΘ = Substitute values, noting that Wf = 2W;. (B) What is the angular speed of the wheel at t = 2.00 s? Solve for the angular displacement and convert to revolutions. w₁t + 12/1at² = (2.00 rad/s)(2.00 s) + (3.50 rad/s²)(2.00 s)² = 11.0 rad A0 (11.0 rad) (1.00 rev/2л rad) = (C) What angular displacement (in revolutions) results during the time in which the angular speed found in part (b) doubles? w/2² -w₁² = 2a40 w = w;+ at = 2.00 rad/s + (3.50 rad/s²) (2.00 s) = 9.00 rad/s = 1.75 rev (2 x 9.00 rad/s)² – (9.00 rad/s)² = 2(3.50 rad/s²)40 A0 (34.7 rad)( = 1 rev 2л rad = 5.52 rev LEARN MORE REMARKS The result of part (b) could also be obtained using the equation ² = w;² + 2a40 and the results of part (a). QUESTION Suppose the radius of the wheel is doubled. Are the answers affected? If so, in what way? (Select all that apply.) The angular speed at t = 2.00 s is the same. The angle rotated through from t = 0 to t = 2.00 s is greater. The angular speed at t = 2.00 s is greater. The angular speed at t = 2.00 s is smaller. The angle rotated through from t = 0 to t = 2.00 s is the same. The angle rotated through from t = 0 to t = 2.00 s is smaller. PRACTICE IT Use the worked example above to help you solve this problem. A wheel rotates with a constant angular acceleration of 3.70 rad/s2. Assume the angular speed of the wheel is 2.25 rad/s at ti = 0. (a) Through what angle does the wheel rotate between t = 0 and t = 2.00 s? Give your answer in radians and revolutions. rad rev (b) What is the angular speed of the wheel at t = 2.00 s? rad/s (c) What angular displacement (in revolutions) results while the angular speed found in part (b) doubles? rev EXERCISE Use the values from PRACTICE IT to help you work this exercise. (a) Find the angle through which the wheel rotates between t = 2.00 s and t = 3.80 s. rad (b) Find the angular speed when t = 3.80 s. rad/s HINTS: GETTING STARTED I I'M STUCK! (c) What is the magnitude of the angular speed five revolutions following t = 3.80 s? rad/s Need Help? Read It
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