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The shaft of a vacuum cleaner motor rotates with an angular acceleration of four times the shaft’s angular velocity raised to the ¾ power. The vacuum beater bar is attached to the motor shaft with pulley through a drive belt. The radii of the motor pulley and the beater bar are 0.25 in and 1.0 in respectively. Determine the angular velocity of the beater bar when t = 4 s, given that omega_0 is 1 rad/s when theta = 0.

Respuesta :

klefenz

Answer:

470 rad/s

Explanation:

The acceleration of the motor shaft is:

γ1 = 4*w1^(3/4)

When connected by a belt the pulleys have the same tangential speed

vt = w * r

vt1 = vt2

w1 * r1 = w2 * r2

w2 = w1 * r1/r2

Therefore:

γ2 = 4 * (w1 * r1/r2)^(3/4)

d(w1 * r1/r2)/dt = 4 * (w1 * r1/r2)^(3/4)

(r1/r2) * dw1/dt = 4 * (r1/r2)^(3/4) * (w1 * r1/r2)^(3/4)

dw1/dt = 4 * (r1/r2)^(-1/4) * (w1)^(3/4)

This is a differential equation.

Solving it through Wolfram Alpha:

w1(t) = (1 / 256) * (4 * (r1/r2)^(-1/4) * t - 4)^4

w1(4) = (1 / 256) * (4 * (0.25 / 1)^(-1/4) * 4 - 4)^4 = 470 rad/s

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