91影视

1. The magnitude of the constant horizontal force applied to a wheel is$F_{app}=2\text{N}$
2. The radius of the wheel, R = 0.3 m
3. The mass of the wheel, m = 10 kg
4. The acceleration of the center of mass of the wheel is,$a=0.60\text{m}/{\text{s}}^2$

The rate of change of angular velocity of the wheel with respect to time is known as angular acceleration. Find the frictional force acting on a wheel by applying Newton鈥檚 second law to the system. Using the Formulae for torque in terms of force and in terms of angular acceleration, we can calculate the moment of inertia of the wheel.

The torque acting on the wheel, is given as-

$$\begin{gathered} \mathbf{蟿}\mathbf{=}\mathbf{F}\mathbf{脳}\mathbf{R} \\ \mathbf{=}\mathbf{滨伪} \end{gathered}$$

Using Newton鈥檚 second law, we get,

$$\begin{gathered} F_{app}-F_f=ma \\ {F}_f=F_{app}-ma \end{gathered}$$

For the given values, the equation becomes-

$$\begin{gathered} F_f=10\text{N}-\left(10\text{kg}\right)\left(0.6m/s^2\right) \\ =0.4\text{N} \end{gathered}$$

The frictional force can be written in the vector notation as

$\stackrel{鈫�}{F_f}=\left(-0.4\text{N}\right)\hat{i}$

The angular acceleration of the wheel is given as,

$$\begin{gathered} 伪=\frac{a}{R} \\ =\frac{0.6m/s^2}{0.3\text{m}} \\ =2rad/s^2 \end{gathered}$$

The torque acting on the wheel is,

$$\begin{gathered} 蟿=F脳R \\ =I伪 \end{gathered}$$

So,

$$\begin{gathered} i伪=F脳R \\ I=\frac{F脳R}{伪} \\ =\frac{0.3\text{N}脳4\text{m}}{2} \\ =0.6{\text{kgm}}^{\text{2}} \end{gathered}$$

The rotational inertia of the wheel about the rotation axis through its center of mass is 0.6 kgm².