A turntable that spins CCW at 78.0 rpm about its center takes 3.5s to reach this

Question

A turntable that spins CCW at 78.0 rpm about its center takes 3.5s to reach this angular speed after it is turned on. Assume the turntable is a solid diskof mass 1.21kg and radius 9.10in

A) Find the angular acceleration magnitude (assumed uniform)

B)What magnitde angular displacement does the turntable make as it speeds up?

C)Determine the net acceleration of a point P on the turntable rim as the turntable reaches.

D)How much work was done on the turntable to accelerate it to 78.0 rpm?

Please show as much work as possible. If its to many questions just answer what you can.

Explanation / Answer

A) First we need to convert 78.0 rpm to standard units:

78 rotations/minute * 2 rad/1 rotation * 1 minute/60 seconds = 2.6 rad/s

From the definition of angular acceleration, we have

= f/t = (2.6 rad/s)/(3.5 s) = 26/35 2.33 rad/s2.

B) The following equation gives angular displacement in terms of initial angular velocity i, angular acceleration , and time t:

= it + 0.5(t2)

Subsituting in i = 0 and = f/t, we get

= 0.5ft  

   = 0.5(2.6 rad/sec)(3.5 s) = 14.3 radians 2.38 rotations.

Not sure about the other parts of the question. Maybe you could be more clear on C?

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