Question

A yo-yo-shaped device mounted on a horizontal frictionless axis is used to lift a$\mathbf{\text{30kg}}$box as shown in Fig$\mathbf{\text{10-59}}$. . The outer radius R of the device is , and the radius r of the hub is$\mathbf{\text{0.20m}}$ . When a constant horizontal force of magnitude $\mathbf{\text{140\hspace{0.17em}N}}$ is applied to a rope wrapped around the outside of the device, the box, which is suspended from a rope wrapped around the hub, has an upward acceleration of magnitude${\mathbf{\text{0.80\hspace{0.17em}m/s}}}^{\mathbf{\text{2}}}$.What is the rotational inertia of the device about its axis of rotation?

Short answer

The rotational inertia of the device about its axis of rotation is$1.6{\text{\hspace{0.17em}kgm}}^2$.

Step-by-step solution

Step 1: Given

i) Mass of box, $m=30\text{\hspace{0.17em}kg}$

ii) Outer radius of the device,$R=0.50\text{\hspace{0.17em}M}$

iii) Inner radius of the device,$r=0.20\text{m}$

iv) Applied force, $F_{app}=140\text{N}$

v) Acceleration, $a=0.80\text{\hspace{0.17em}m}/{\text{s}}^2$

Determining the concept

Find the tension in the rope using Newton’s second law of motion. This tension can be used to find the moment of inertia of the device about its axis of rotation by applying the equation of the rotational motion to the device.

Formulae are as follow:

$T-mg=ma$

$F_{app}R-Tr=I\alpha$

$\alpha =a/r$

where,$F_{app}$ is apparent force, m is mass, R, r are radii, I is moment of inertia, T is tension, a is an acceleration,g is an acceleration due to gravity and $\alpha$ is angular acceleration.

Determining therotational inertia of the device about its axis rotation 

For the tension in the rope,

$T-mg=ma$

$\begin{array}{c}T=ma+mg\\ =(30\text{kg})\left(0.80\frac{\text{m}}{{\text{s}}^{\text{2}}}\right)+(30\text{\hspace{0.17em}kg})\left(9.8\frac{\text{m}}{{\text{s}}^{\text{2}}}\right)\\ =318\text{\hspace{0.17em}N}\end{array}$

Now, an equation for the rotational motion to the device is,

$F_{app}R-Tr=I\alpha$

As,

$\alpha =a/r$

$\begin{array}{l}{F}_{ap}R-Tr=I\left(\frac{a}{r}\right)\\ I=\frac{(F_{ap}R-Tr)}{\left(\frac{a}{r}\right)}\\ I=(F_{ap}R-Tr)\left(\frac{r}{a}\right)\end{array}$

For the given values, the above equation becomes-

$\begin{array}{c}I=(140\text{\hspace{0.17em}N}\left(0.50\text{\hspace{0.17em}m}\right)-\left(318\text{\hspace{0.17em}N}\right)0.20\text{\hspace{0.17em}m})\left(\frac{0.20\text{\hspace{0.17em}m}}{0.8\text{\hspace{0.17em}m}/{\text{s}}^2}\right)\\ =1.6\text{\hspace{0.17em}kg}.{\text{m}}^2\end{array}$

Hence, the rotational inertia of the device about its axis rotation is$1.6{\text{\hspace{0.17em}kgm}}^2$.