##### Physics: Mechanical Energy

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A horizontal 810-N merry-go-round of radius 1.70 m is started from rest by a constant horizontal force of 45 N applied tangentially to the merry-go-round. Find the kinetic energy of the merry-go-round after 4.0 s.

(Assume it is a solid cylinder. Also assume the force is applied at the outside edge.)

Oct 16th, 2017

Mass of the merry go round =810/9.8 =82.65 kg

F=ma

a=F/m

v/t=F/m

v=Ft/m

v=45*4/82.65

=2.17 m/s

Linear Kinetic energy =1/2 m v^2

=1/2 * 82.65* 2.17^2

=194.22  Joules

let us find angular or rotational kinetic energy

I =1/2 mR^2

=1/2 *82.65 * 1.70^2

=119.42 kg m^2

Torque =F.r

=45*1.70

=76.5 =Ia

angular velocity w=T *t/I

=76.5*4/119.42

Rotational kinetic energy =1/2 I w^2

=1/2 *119.42 *2.56^2

=391.3 J

T.K.E =391.3+194.22

=585.52 J

Jul 5th, 2015

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Oct 16th, 2017
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Oct 16th, 2017
Oct 17th, 2017
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