Question

Figure $\mathbf{10}\mathbf{-}\mathbf{54}$shows a flat construction of two circular rings that have a common center and are held together by three rods of negligible mass. The construction, which is initially at rest, can rotate around the common center (like a merry-go-round), where another rod of negligible mass lies. The mass, inner radius, and outer radius of the rings are given in the following table. A tangential force of magnitude $12.0N$is applied to the outer edge of the outer ring for $0.300s$.What is the change in the angular speed of the construction during the time interval?

Ring	Mass (kg)	Inner Radius (m)	Outer Radius (m )
1	0.120	0.0160	0.0450
2	0.24	0.0900	0.1400

Short answer

Change in angular speed of construction during the time interval is$146\raisebox{1ex}{$\text{rad}$}\!\left/ \!\raisebox{-1ex}{$\text{sec}$}\right.$

Step-by-step solution

Given

1. Mass of ring 1 is$0.120\text{kg}$
2. Mass of ring 2 is$0.24\text{kg}$
3. Inner radius of ring 1 is$0.016\text{m}$
4. Outer radius of ring 1 is$0.0450\text{m}$
5. Inner radius of ring 2 is$0.09\text{m}$
6. Outer radius of ring 2 is$0.14\text{m}$
7. Magnitude of tangential force is$12.0\text{N}$

To understand the concept

Use the formula in terms of force and radius to find the torque. Using another formula for torque in terms of rotational inertia and angular acceleration, find the angular acceleration.

Formula:

$\tau =\text{r}\times \text{F}$

$\tau =\text{I}\alpha$

Calculate the moment of inertia of the two rings

For ring,the inertiais l,

${\text{I}}_1=\frac{1}{2}{\text{M}}_1\left({\text{R}}_{1i}^2+{\text{R}}_{1o}^2\right)$

is the mass of the ring, and ${\text{R}}_{1i}$and ${\text{R}}_{1o}$is the innerand outer radius of the smaller ring respectively.

$$\begin{gathered} {\text{I}}_1=\frac{1}{2}\times 0.120\times \left({0.0160}^2+{0.045}^2\right) \\ {\text{I}}_1=1.3686\times {10}^{-4}\text{kg}-{\text{m}}^2 \end{gathered}$$

For ring 2,

$$\begin{gathered} {\text{I}}_2=\frac{1}{2}{\text{M}}_2\left({\text{R}}_{2i}^2+{\text{R}}_{2o}^2\right) \\ {\text{I}}_2=\frac{1}{2}\times 0.24\times \left({0.090}^2+{0.14}^2\right) \\ {\text{I}}_2=3.324\times {10}^{-3}\text{kg}-{\text{m}}^2 \end{gathered}$$

Calculate the total inertia

$M_2$is mass of the ring and ${\text{R}}_{2i}$and ${\text{R}}_{2o}$is the inner and outer radius of the larger ring respectively.

Total inertia is as follows:

$$\begin{gathered} \text{I}={\text{I}}_1+{\text{I}}_2 \\ \text{I}=1.3686\times {10}^{-4}+3.324\times {10}^{-3} \\ \text{I}=3.460\times {10}^{-3}\text{kg}-{\text{m}}^2 \end{gathered}$$

Calculate the torque

Now, torque as follows:

$$\begin{gathered} \tau ={\text{R}}_{2o}\times \text{F} \\ \tau =0.14\times 12 \\ \tau =1.68\text{N}.\text{m} \end{gathered}$$

Calculate the angular acceleration

We know

$$\begin{gathered} \tau =\text{I}\alpha \\ 1.68=3.46\times {10}^{-3}\times \alpha \end{gathered}$$

So

$\alpha =485\text{rad}/\text{sec}$

Calculate the angular speed

Now, angular speed is as follows:

$$\begin{gathered} \omega =\alpha \times \text{t} \\ \omega =485\times 0.3 \\ \omega =145.5\text{rad}/\text{sec} \\ \omega =146\text{rad}/\text{sec} \end{gathered}$$