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GaryK [48]
3 years ago
7

An electric ceiling fan is rotating about a fixed axis with an initial angular velocity magnitude of 0.270 rev/s . The magnitude

of the angular acceleration is 0.895 rev/s^2 . Both the the angular velocity and angular accleration are directed clockwise. The electric ceiling fan blades form a circle of diameter 0.790 m .
1. Compute the fan's angular velocity magnitude after time 0.194s has passed. (rev/s)
2. Through how many revolutions has the blade turned in the time interval 0.194s from Part A? (rev)
3. What is the tangential speed vtan(t) of a point on the tip of the blade at time t = 0.194s ? (m/s)
4. What is the magnitude a of the resultant acceleration of a point on the tip of the blade at time t = 0.194s ? (m/s^2)
Physics
1 answer:
JulijaS [17]3 years ago
5 0

Answer:

Given information is summarized as below:

\omega _{o}=0.270rev/sec=1.69 rad/sec\\\alpha =0.895rev/sec=5.62rad/sec\\d=0.790m

1)Using first equation of kinematics for angular motion we have

\omega _{f}=\omega _{o}+\alpha t\\\\\therefore \omega _{f}=1.69+5.62\times 0.194\\\omega _{f}=2.780rad/sec

2) Using second equation of kinematics for angular motion we have

\theta =\omega _{o}t+\frac{1}{2}\alpha t^{2}\\\\\theta =1.69\times 0.194+05\times 5.62\times0.194^{2}\\\\\theta = 0.4336rad

3) The tangential speed is given as

v_{t}=\omega _{f}\times r\\\\v_{t}=2.78\times\frac{0.790}{2}=1.098m/s

4)

The resultant acceleration is given by

a_{res}=\sqrt{(a_{tangential})^{2}+(a_{radial})^{2}}\\\\a_{res}=\sqrt{(\alpha r)^{2}+(\omega _{f}^{2}r)^{2}}\\\\a_{res}=3.77m/s^{2}

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