Q11P - Fundamentals Of Physics

Question

Question: In Figure, a constant horizontal force${\overrightarrow{\mathbf{F}}}_{\mathbf{a}\mathbf{p}\mathbf{p}}$ of magnitude is applied to a wheel of mass 10 kg   and radius0.30 m . The wheel rolls smoothly on the horizontal surface, and the acceleration of its centre of mass has magnitude 0.60 m/s²  (a) In unit-vector notation, what is the frictional force on the wheel? (b) What is the rotational inertia of the wheel about the rotation axis through its centre of mass? 

Step-by-Step Solution

Answer

Answer

 

1. The frictional force acting on a wheel in the unit-vector notation$\left(-4.0 \text{N}\right)\hat{i}$.
2. The rotational inertia of the wheel about the rotation axis through its center of mass is$0.6{\text{kgm}}^2$ .

Step 1: Given

 

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,$a=0.60\text{m}/{\text{s}}^2$

Step 2: To understand the concept

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’s 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-

$$\begin{gathered} \mathbf{\tau }\mathbf{=}\mathbf{F}\mathbf{\times }\mathbf{R} \\ \mathbf{=}\mathbf{I\alpha } \end{gathered}$$

Step 3: (a) Calculate the frictional force on the wheel in unit-vector notation

Using Newton’s 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-

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

The frictional force can be written in the vector notation as

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

Step 4: (b)

The angular acceleration of the wheel is given as,

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

The torque acting on the wheel is,

$$\begin{gathered} \tau =F\times R \\ =I\alpha \end{gathered}$$

So,

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

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