COLLEGE PHYSICS-ACHIEVE AC (1-TERM)
COLLEGE PHYSICS-ACHIEVE AC (1-TERM)
3rd Edition
ISBN: 9781319453916
Author: Freedman
Publisher: MAC HIGHER
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Chapter 8, Problem 102QAP
To determine

Magnitude and direction of the torque.

Expert Solution & Answer
Check Mark

Answer to Problem 102QAP

  G=325.83Nm

Direction = clockwise

Explanation of Solution

Given info:

Weight of the merry-go-round =m1=325kg

Initial angular speed of dough before applies friction to outer rim =ωbefore=4.70rad/s

Radius of merry-go-round =r0=1.40m

Weight of the child =m2=36kg

Child's distance from rotational axis =r1=1.25m

Time which merry-go-round and the child spent before stopped. =5s

Formula used:

Angular momentum,

  Lz=Iωz ( Lz= Angular momentum, I= moment of inertia, ωz= angular velocity)

Also, it can be written as,

  Lz=τzt ( Lz= Angular momentum, τz

  = torque, t= time)

Conservation of angular momentum,

  Lbefore=Lafter

  Ibeforeωbefore=Iafterωafter

Calculation:

Let's name the torque which applies to outer rim as τ.

Angular momentum of the given torque,

  Lz=τzt...(1)

Substituting the time value,

  Lz=τz(5s)

Let's consider the angular momentum generated by merry-go-round and child before stopped.

Moment of inertia of the merry-go-round before the application of torque.

  Imerrygoround=12m1(r0)2

Angular momentum of the merry-go-round before the application of torque.

  Lmerrygoround=Iω

  Lmerrygoround=12m1(ro)2ωbefore...(2)

Moment of inertia of the child before the application of torque.

  Ichild=12m2(r1)2

Angular momentum of the merry-go-round before the application of torque.

  Lchild=Iω

  Lchild=12m2(r1)2ωbefore...(3)

According to the Conservation of angular momentum,

  Lbefore=Lafter

  Lafter= Angular momentum caused by the torque

  Lmerrygoround+Lchild=τt

Substituting (1),(2),(3) expressions,

  12m1(ro)2ωbefore+12m2(r1)2ωbefore=τt

  12m1(ro)2ωbefore+12m2(r1)2ωbefore=τ(5s)

  12ωbefore[m1(ro)2+m2(r1)2]=τ(5s)

Substituting the given values, we get

  12(4.70rad/s)[325kg(1.40m)2+36kg(1.25m)2]=τ(5s)

  12(4.70rad/s)[637kgm2+56.25kgm2]=τ(5s)

  12(4.70rad/s)[693.25kgm2]=τ(5s)

  (4.70rad/s)[693.25kgm2]2(5s)=τ

  τ=3258.275rad/skgm210s

  τ=325.8275Nm

  τ325.83Nm

  τ=325.83Nm

And direction of torque should be clockwise direction, because the directions of angular momentum of merry-go-round and child are counter-clockwise.

Conclusion:

Thus, magnitude of the torque is 325.83Nm and direction of the torque is clockwise

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Chapter 8 Solutions

COLLEGE PHYSICS-ACHIEVE AC (1-TERM)

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