Question

An old record completes 0.890 revolutions as it slows down
uniformly from 78.0 rpm to 22.8 rpm. What is the magnitude of the
angular acceleration of the record, in rad/s^{2}?

Answer #1

Under uniform acceleration,(wf)^2=(wi)^2+2*a*s, where wf is final angular velocity,wi is initial angular velocity ,a is angular acceleration and s is angular displacement.

Here,wf=22.8 rpm=22.8*2π/60 rad/s=2.3876 rad/s, wi=78 rpm=78*2π/60=8.16814 rad/s, s=0.890 revolutions=0.890*2π radians=5.592 radians.

So, (2.3876)^2=(8.16814)^2+2*a*5.592

=>5.7=66.72+11.184a

=>a=(5.7-66.72)/11.184= -5.456 rad/s2.

So, magnitude of angular acceleration is 5.456 rad/s2 and the negative sign denotes that the object is decelerating.

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shut off and uniformly slowed down and stopped in 5.5 seconds.
(a) What was the magnitude of its angular acceleration (in
rad/s2) as it slowed down?
(b) Through how many revolutions did it turn while stopping?

When an old LP turntable was revolving at 33(1/3)rpm, it was
shut off and uniformly slowed down and stopped in 5.5 seconds.
a) What was the magnitude of its angular acceleration (in
rad/s^2) as it slowed down?
b) Through how many revolutions did it turn while stopping?

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s .
a) Calculate its angular acceleration, assumed constant.
b) Calculate the total number of revolutions the engine makes in
this time.

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down and stops in 30 sec after the motor is turned off. (a) Find
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make in this time?

An automobile engine slows down from 3900 rpmto 1500 rpm in 2.00
s .
Calculate its angular acceleration, assumed constant.
Calculate the total number of revolutions the engine makes in
this time.

A rolling wheel of radius 34 cm slows down uniformly from 8.4
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What is the (constant) angular acceleration of the
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α=α= rad/s2
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t=t= s
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KErot = J

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