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).
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).
Show Hint
Correct Answer: 1.3 - 1.4
Solution and Explanation
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:
Itotal = 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 = Itotal × 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)
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