A flywheel starts from rest, and has an angular acceleration of 4 rad/s2. What will be its angular speed when it has made one revolution?

? 04/15/2018. 6 answers
Science & Mathematics Physics

6 Answers


electron1 04/16/2018.

One revolution is 2 π radians. Let’s use the following equation to solve this problem.

ωf^2 = ωi^2 + 2 * α * θ, ωi = 0

ωf^2 = 2 * 4 * 2 * π

ω = 4 * √π

To one significant digit, this is approximately 7 rad/s.


bramhananda r 04/16/2018.

From equations of motion

w^2 - w0^2 = 2*alpha*theta

theta = 2pi

W^2 = 2*4*2pi ==>w = sqrt(16pi) = 4 sqrt(pi) rad/s = 7.089 rad/s


Andrew Smith 04/16/2018.

You can use all your known linear formulae on circular motion.

v^2 = 2as

v = sqrt( 2as) = sqrt( 2 * 4 * 2 * pi() ) ( the "distance" moved is 2 pi() radians )

= 7.09 radians per second.

Why learn a whole heap of "new" formulae if you already know them all anyway.


donpat 04/16/2018.

You have :

-----------------------

a = dw/dt --------> dt = dw/a

w = d rad / dt ----> dt = d rad / w

dt = dt = dw/a = d rad / w

w dw = a d rad/dt

Integral [ w dw ] w1--->w2 = Integral [ a drad ] rad1--->rad2

( w2^2 - w1^2 ) / ( 2 ) = ( a ) ( rad2 - rad1 )

w2 = SQRT [ ( 2 ) ( a ) ( 2 pi rad ) ] = SQRT [ (4 ) ( pi ) ( a ) ]

w2 = SQRT [( 2) ( 2 pi rad/rev ) ( 4 rad/s^2 )]

w2 = SQRT [ ( 16 ) ( pi rad^2/s^2 ) ]

w2 = 7.1 rad/s <--------------------------------------


Steve4Physics 04/16/2018.

For linear acceleration we have vf² = vi² + 2as.

The equivalent for rotational acceleration is ωf² = ωi² + 2αθ

Since one rotation is an angular displacement of θ = 2π:

ωf² = 0² + 2*4*2π

. . . = 50.27

ωf = 7.09 rad/s


? 04/16/2018.

2pi = 1/2 * a * time^2

find the time

angular speed = angular acceleration * time

7,1 rad/sec


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