The wheels and axle system lying on a rough surface is shown in the figure.

Each wheel has diameter 0.8 m and mass 1 kg. Assume that the mass of the wheel is concentrated at rim and neglect the mass of the spokes. The diameter of axle is 0.2 m and its mass is 1.5 kg. Neglect the moment of inertia of the axle and assume g = 9.8 m/s² . An effort of 10 N is applied on the axle in the horizontal direction shown at mid span of the axle. Assume that the wheels move on a horizontal surface without slip. The acceleration of the wheel axle system in horizontal direction is ______________𝐦/𝐬² (round off to one decimal place).

The correct answer is 1.36.

1. Understanding the System Geometry and Given Parameters:
Based on the first image, the system consists of two wheels connected by a central axle lying on a rough horizontal surface. The labeled parameters are:
• Diameter of each wheel = 0.8 m, which gives a wheel radius of $R=0.4\text{m}$.
• Mass of each wheel = 1 kg, concentrated entirely at the rim (modeled as a thin hoop).
• Diameter of the axle = 0.2 m, which gives an axle radius of $r=0.1\text{m}$.
• Mass of the axle = 1.5 kg, with its own moment of inertia about its center axis neglected.
• A horizontal force of 10 N is applied at the mid-span of the axle. As shown by the arrow and dimensions in the second diagram, the line of action of this 10 N force is along the bottom of the axle.

2. Locating the Instantaneous Center of Rotation:
Since the wheels roll on the horizontal surface without slipping, the contact point between the wheels and the ground acts as the instantaneous center of rotation, denoted as point $I$.
• The center of the axle is at a height of $R=0.4\text{m}$ above the ground.
• The line of action of the 10 N force is along the bottom surface of the axle, which lies at a distance of $r=0.1\text{m}$ below the center axis.
• Therefore, the perpendicular distance from the contact point $I$ to the line of action of the 10 N force is:
$h=R-r=0.4\text{m}-0.1\text{m}=0.3\text{m}$

3. Calculating the Total Moment of Inertia about the Instantaneous Center (I):
We apply the parallel axis theorem to find the moment of inertia of both the wheels and the axle about the ground contact point $I$:
• For the wheels:
Each wheel of mass $m_w=1\text{kg}$ behaves as a thin cylindrical shell (hoop), so its moment of inertia about its center of mass is:
$I_{G_w}=m_w{R}^2=1\times {0.4}^2=0.16{\text{kg·m}}^2$
Using the parallel axis theorem, the moment of inertia of a single wheel about point $I$ is:
$I_{I_w}=I_{G_w}+m_w{R}^2=0.16+1\times {0.4}^2=0.32{\text{kg·m}}^2$
For both wheels combined, the moment of inertia about point $I$ is:
$2\times I_{I_w}=2\times 0.32=0.64{\text{kg·m}}^2$

• For the axle:
The axle has mass $m_a=1.5\text{kg}$. Neglecting its own moment of inertia about its axis ($I_{{\mathrm{axle}}_o}=0$), we only account for the displacement of its mass center to the contact point $I$ at distance $R$:
$I_{{\mathrm{axle}}_I}=I_{{\mathrm{axle}}_o}+m_a{R}^2=0+1.5\times {0.4}^2=0.24{\text{kg·m}}^2$

• Total moment of inertia of the entire assembly about the instantaneous center $I$ is:
$I_I=0.64+0.24=0.88{\text{kg·m}}^2$

4. Setting up the Equation of Motion:
The net torque about the instantaneous center of rotation $I$ is given by:
$\sum T_I=F\times h=10\text{N}\times 0.3\text{m}=3\text{N·m}$

Using the rotational form of Newton's second law about point $I$:
$\sum T_I=I_I\alpha$
$3=0.88\alpha$
$\alpha =\frac{3}{0.88}\approx 3.4091{\text{rad/s}}^2$

5. Finding the Linear Acceleration:
The linear acceleration $a$ of the center of the wheels and axle system is directly related to the angular acceleration $\alpha$ by:
$a=\alpha R=3.4091\times 0.4\approx 1.36{\text{m/s}}^2$

Rounding to one decimal place, the horizontal acceleration of the wheel-axle system is 1.4 m/s² (or exactly 1.36 m/s² as specified in the options).