Question

A wheel of radius $\mathbf{0}\mathbf{.}\mathbf{20}\mathbf{}\mathit{m}\mathbf{}$is mounted on a frictionless horizontal axis. The rotational inertia of the wheel about the axis is$\mathbf{0}\mathbf{.}\mathbf{050}\mathit{k}\mathit{g}\mathbf{.}{\mathit{m}}^{\mathbf{2}}$ . A mass less cord wrapped around the wheel is attached to a $\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{}\mathit{k}\mathit{g}$ block that slides on a horizontal frictionless surface. If a horizontal force of magnitude $\mathit{P}\mathbf{=}\mathbf{}\mathbf{3}\mathbf{.}\mathbf{0}\mathit{N}$ is applied to the block as shown in Fig.$\mathbf{10}\mathbf{-}\mathbf{56}$, what is the magnitude of the angular acceleration of the wheel? Assume the cord does not slip on the wheel.

Short answer

The magnitude of the angular acceleration of the wheel is $4.62{\text{rad/s}}^{\text{2}}\text{.}$

Step-by-step solution

Step 1: Given

1. Mass of block, $M=2.0\text{kg}$
2. Radius of wheel, $R=0.20\text{m}$
3. Rotational inertia of wheel, $I=0.050{\text{kgm}}^2$
4. Magnitude of force, $P=3.0\text{N}$

Determining the concept

Due to applied force$\overrightarrow{\mathbf{P}}$, the block will accelerate. So, using Newton’s second law, write a net force equation for the block. Similarly, use Newton’s second law for rotating the wheel. Then by using the relation between angular acceleration and linear acceleration, find the magnitude of the angular acceleration of the wheel.

Formulae are as follow:

$$\begin{gathered} P-T=ma \\ -TR=I\alpha \\ {a}_t=R\alpha \end{gathered}$$

where, T, P are forces, m is mass, R is radius, I is moment of inertia, a is acceleration and is angular acceleration.

Determining the magnitude of the angular acceleration of the wheel

Taking rightward motion to be positive for the block and clockwise motion to be negative for the wheel.

Applying Newton’s second law to the block gives,

$P-T=ma ---\left(1\right)$

Similarly, applying Newton’s second law to the wheel gives,

$-TR=I\alpha$

Now,

$a_t=R\alpha$

As tangential acceleration $a_t$is opposite to that of block’s acceleration ,

$$\begin{gathered} a_t=-a \\ -a=R\alpha \\ \alpha =-\frac{a}{R} \end{gathered}$$

Therefore,

$$\begin{gathered} -TR=-I\frac{a}{R} \\ T=\frac{Ia}{R^2} ---\left(2\right) \end{gathered}$$

Use equation (2) into (1),

$$\begin{gathered} P-\frac{Ia}{R^2}=ma \\ P-\frac{Ia}{R^2}=-m\alpha \\ P=-\left(m\alpha R+\frac{I\alpha R}{R^2}\right) \\ P=-\alpha R\left(m+\frac{I}{R^2}\right) \\ \alpha =-\frac{P}{R\left(m+\frac{I}{R^2}\right)} \\ \alpha =-\frac{3.0\text{N}}{\left(0.20\text{m}\right)\left[\left(2.0\text{kg}\right)+\frac{(0.0050\text{kg}.{\text{m}}^2}{{\left(0.20\text{m}\right)}^2}\right]} \\ \alpha =-4.62\raisebox{1ex}{$rad$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right. \end{gathered}$$

Hence, the magnitude of the angular acceleration of the wheel is$4.62\text{rad}/{\text{s}}^2$

Therefore, the magnitude of the angular acceleration of the wheel can be found using Newton’s second law for the block and for the rotating wheel by using the relation between angular acceleration and linear acceleration.