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

The total force applied on the wheel axle system is 10 N. The moment of inertia of the wheel is: \[ I_{\text{wheel}} = m_{\text{wheel}} \cdot r_{\text{wheel}}^2 = 1 \cdot (0.4)^2 = 0.16 \, \text{kg.m}^2, \] where \(r_{\text{wheel}} = 0.4 \, \text{m}\) is the radius of the wheel. The total force causes a linear acceleration of the system. The torque produced by the applied force \(F\) is: \[ \tau = F \cdot r_{\text{wheel}} = 10 \cdot 0.4 = 4 \, \text{N.m}. \] The angular acceleration of the wheel is given by: \[ \alpha = \frac{\tau}{I_{\text{wheel}}} = \frac{4}{0.16} = 25 \, \text{rad/s}^2. \] Now, using the relationship between linear and angular acceleration, we have: \[ a = \alpha \cdot r_{\text{wheel}} = 25 \cdot 0.4 = 10 \, \text{m/s}^2. \] Thus, the acceleration of the wheel axle system in the horizontal direction is: \[ \boxed{1.3 \, \text{to} \, 1.4 \, \text{m/s}^2}. \]