Chapter 8.2, Problem 30PP

To determine

The force required to provide the system a given angular acceleration.

Answer to Problem 30PP

The force required to provide the system an angular acceleration of $2.57{\text{rad/s}}^2$ is $7.7\text{N}$ .

Explanation of Solution

Given:

The moment of inertia of the larger disk is $I_{\text{L}}=0.26\text{kg}\cdot {\text{m}}^2$ .

The diameter of the smaller disk is $d=0.180\text{m}$ .

The mass of the smaller disk is $m=2.5\text{kg}$ .

The angular acceleration of the system is $\alpha =2.57{\text{rad/s}}^2$ .

Formula used:

The expression for the moment of inertia $\left(I\right)$ of a circular disk is

  $I=\frac{1}{2}m{r}^2$

Here, $m$ is the mass of the disk and $r$ is the distance from the axis of rotation.

The expression for the angular acceleration $\left(\alpha \right)$ of the disk is

  $\alpha =\frac{Fr}{I}$

Here, $I$ is the moment of inertia of the system, $F$ is the force acting on the system, and $r$ is the radius of the disk.

Calculation:

The radius of the smaller disk $\left(r\right)$ is,

  $\begin{array}{l}r=\frac{d}{2}\\ r=\frac{0.180\text{m}}{2}\\ r=0.090\text{m}\end{array}$

The moment of inertia $\left(I_{\text{S}}\right)$ of the smaller disk is

  $\begin{array}{l}{I}_{\text{S}}=\frac{1}{2}m{r}^2\\ I_{\text{S}}=\frac{1}{2}\times 2.5\text{kg}\times {\left(0.090\text{m}\right)}^2\\ I_{\text{S}}=0.01\text{kg}\cdot {\text{m}}^2\end{array}$

So, the force $\left(F\right)$ acting on the disk is

  $\begin{array}{l}F=\frac{\alpha \left(I_{\text{S}}+I_{\text{L}}\right)}{r}\\ F=\frac{2.57{\text{rad/s}}^2\times \left(0.01\text{kg}\cdot {\text{m}}^2+0.26\text{kg}\cdot {\text{m}}^2\right)}{0.090\text{m}}\\ F=7.7\text{N}\end{array}$

Conclusion:

Therefore, the force required to provide the system an angular acceleration of $2.57{\text{rad/s}}^2$ is $7.7\text{N}$ .