Question

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.

Answer 1

The angular velocity of the beater bar at t=4s is approximately 6.37 rad/s, based on the given angular acceleration equation and initial angular velocity.

Step-by-step explanation:

The angular velocity of the beater bar can be found using the relationship between angular acceleration and angular velocity. The given equation states that the angular acceleration is four times the angular velocity raised to the 3/4 power. Therefore, we can write:

α = 4 * ω^(3/4)

To find the angular velocity at t=4s, we can integrate the equation to get:

ω = 4/7 * t^7/4 + C

When t = 0, ω = ω_0 = 1 rad/s. Substituting these values, we can solve for C:

1 = 0 + C

Therefore, C = 1. Finally, we can substitute t = 4s into the equation to get the angular velocity:

ω = 4/7 * 4^7/4 + 1

Calculating this expression, we find that the angular velocity of the beater bar at t=4s is approximately 6.37 rad/s.

Answer 2

470 rad/s

Step-by-step 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