Alpha Wheeled


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Alpha Wheeled
physics please help ….. im confused …….. thank ID 10 pts 4 d best answer too … thank u in advance.?

As a result of friction, the angular velocity of a wheel changes with time among the following, d (theta) / dt = w0e ^ – (alpha) (t), where w0 and alpha are constants. The changes in angular velocity of 4.30 rad / s at t = 0 to 2.00 rad / s at t = 9.50 s. Use this information to determine alpha and w0. (A) Use alpha and w0, calculate the magnitude angular acceleration at t = 3.00 s. rad/s2 (b) Use of alpha and W0, determine the number of revolutions the wheel makes in the first 2.50 s. rev (c) Use of alpha and W0, determine the number of revolutions made before reaching the break. rev

Let w the angular velocity: w = d (theta) / dt = ° we ^ – (a) at t = 0, w (0) = ° w = 4.30 rad / s. At t = 9.50 s, w (9,5) = ° we ^ – (a * 9.5) = 2.00 rad / se ^ – (a * 9.5) = 2/4.3 —> – (a * 9.5) = ln (2/4.3) a = ln (2.15) / 9.5 = 0.080 s ^ -1. (A) The angular acceleration is dw / dt = – AW ° e ^ – (a) at t = 3 s, dw / dt = – 0.080 * 4.30 * E ^ – (0,080 * 3) = – 0.272 rad / s ^ 2 (b) theta (t) = (° w / a) (1 – e ^ – (a)) at t = 2.5 s, theta n = 9.736 rad = theta / 2 pi = 1.55 rev. (c) w (t) = 0 e ^ – (a) = 0 then ° theta = w / A and N = theta / 2 pi = 8.49 rev.

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