Chapter 8, Problem 102QAP

To determine

Magnitude and direction of the torque.

Answer to Problem 102QAP

  $G=325.83Nm$

Direction $=$ clockwise

Explanation of Solution

Given info:

Weight of the merry-go-round $=m_1=325kg$

Initial angular speed of dough before applies friction to outer rim $={\omega }_{before}=4.70rad/s$

Radius of merry-go-round $=r_0=1.40m$

Weight of the child $=m_2=36kg$

Child's distance from rotational axis $=r_1=1.25m$

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

Formula used:

Angular momentum,

  $L_z=I{\omega }_z$ ( $L_z=$ Angular momentum, $I=$ moment of inertia, ${\omega }_z=$ angular velocity)

Also, it can be written as,

  $L_z={\tau }_{z}t$ ( $L_z=$ Angular momentum, ${\tau }_z$

  $=$ torque, $t=$ time)

Conservation of angular momentum,

  $L_{before}=L_{after}$

  $I_{before}{\omega }_{before}=I_{after}{\omega }_{after}$

Calculation:

Let's name the torque which applies to outer rim as $\tau$.

Angular momentum of the given torque,

  $L_z={\tau }_{z}t$$\mathrm{...}(1)$

Substituting the time value,

  $L_z={\tau }_z(5s)$

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.

  $I_{merry-go-round}=\frac{1}{2}{m}_1{\left(r_0\right)}^2$

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

  $L_{merry-go-round}=I\omega$

  $L_{merry-go-round}=\frac{1}{2}{m}_1{\left(r_o\right)}^2{\omega }_{before}$$\mathrm{...}(2)$

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

  $I_{child}=\frac{1}{2}{m}_2{\left(r_1\right)}^2$

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

  $L_{child}=I\omega$

  $L_{child}=\frac{1}{2}{m}_2{\left(r_1\right)}^2{\omega }_{before}$$\mathrm{...}(3)$

According to the Conservation of angular momentum,

  $L_{before}=L_{after}$

  $L_{after}=$ Angular momentum caused by the torque

  $L_{merry-go-round}+L_{child}=\tau t$

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

  $\frac{1}{2}{m}_1{\left(r_o\right)}^2{\omega }_{before}+\frac{1}{2}{m}_2{\left(r_1\right)}^2{\omega }_{before}=\tau t$

  $\frac{1}{2}{m}_1{\left(r_o\right)}^2{\omega }_{before}+\frac{1}{2}{m}_2{\left(r_1\right)}^2{\omega }_{before}=\tau (5s)$

  $\frac{1}{2}{\omega }_{before}[m_1{(r_o)}^2+m_2{(r_1)}^2]=\tau (5s)$

Substituting the given values, we get

  $\frac{1}{2}(4.70rad/s)[325kg{(1.40m)}^2+36kg{(1.25m)}^2]=\tau (5s)$

  $\frac{1}{2}(4.70rad/s)[637kg{m}^2+56.25kg{m}^2]=\tau (5s)$

  $\frac{1}{2}(4.70rad/s)[693.25kg{m}^2]=\tau (5s)$

  $\frac{(4.70rad/s)[693.25kg{m}^2]}{2(5s)}=\tau$

  $\tau =\frac{3258.275rad/skg{m}^2}{10s}$

  $\tau =325.8275Nm$

  $\tau \approx 325.83Nm$

  $\tau =325.83Nm$

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