Question

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 \( g = 9.8 \, \text{m/s}^2 \). 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 \(\underline{\hspace{1cm}}\) m/s² (round off to one decimal place).

Hint:

The acceleration of the wheel axle system can be determined by considering the rotational inertia of the wheel and the applied force.

Answer 1

Correct Answer: 1.3 - 1.4

Step 1: Total mass of the system

Total mass = (2 × mass of wheel) + mass of axle
Total mass = (2 × 1) + 1.5 = 3.5 kg

Step 2: Translational motion of the system

Let friction force on each wheel be F (opposite to motion).

Net horizontal force on the system:

10 − 2F = 3.5a    ...(1)

Step 3: Rotational motion of the wheels

Moment of inertia of one wheel:

I = mR² = 1 × (0.4)²

Moment of inertia of two wheels:

I_total = 2 × 1 × (0.4)² = 0.32 kg m²

Torque due to friction on both wheels = 2F × 0.4

Torque due to applied force on axle = 10 × 0.1 = 1 N m

Net torque equation:

2F × 0.4 − 1 = I_total × α

For rolling without slipping:

α = a / R = a / 0.4

Substituting:

0.8F − 1 = 0.32 × (a / 0.4)

0.8F − 1 = 0.8a

F − 1.25 = a    ...(2)

Step 4: Solving equations (1) and (2)

From equation (2):
F = a + 1.25

Substitute into equation (1):

10 − 2(a + 1.25) = 3.5a

7.5 = 5.5a

a = 1.36 m/s²

Final Answer:

Acceleration of the wheel–axle system = 1.4 m/s² (rounded to one decimal place)