An electric turntable 0.800 m in diameter is rotating about a fixed axis with an

Question

An electric turntable 0.800 m in diameter is rotating about a fixed axis with an initial angular velocity of 0.290 rev/s . The angular acceleration is 0.914 rev/s2 .

Part A

Compute the angular velocity after a time of 0.199 s .

Part B

Through how many revolutions has the blade turned in this time interval?

Part C

What is the tangential speed of a point on the tip of the blade at time t = 0.199 s ?

Part D

What is the magnitude of the resultant acceleration of a point on the tip of the blade at time t = 0.199 s ?

Explanation / Answer

A)

Radius of fan blade = r = 0.4 m

The initial angular velocity of fan = wi = 0.29 rev/s = 0.29 * 2pi radian/s.

The angular acceleration = a! = 0.914 rev/s^2.= 0.914 * 2pi radian/s^2

time 't'=0.199 s

angular velocity after 0.199 s=wf=?

wf = wi + a!t

wf =0. 29 * 2pi + 0.914 * 2pi * 0.199

wf = 2.963 rad/s

wf = 0.4718 rev./s

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B) number of revolutions=N=angular displacement O/2pi

angular displacement O=[(wf+wi)/2 ]t = [(0.29 + 0.4718 )/2]*2pi*0.199 = 0.0757*2pi radian

number of revolutions=N=O/2pi = 0.0757

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C) tangential speed v = r wf = 0.4 * 2.936 = 1.1852 m/s
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D) resultant acceleration is vector sum of centripetal acceleration and tangential acceleration

centripetal acceleration= Ac = r(wf)^2 = 0.4 * 2.936 * 2.936 = 3.44 m/s^2

tangential acceleration= At = r*a! = 0.4 * 5.74 = 2.296 m/s^2

resultant acceleration = sqrt [Ac^2 + At^2] = sqrt 17.8265 = 4.135 m/s^2

resultant acceleration of a point on the tip at 0.199 s is 4.135 m/s^2

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