Question 36: Rolling Cart on an Incline
Figure 1: Forces acting on the cart (gravity and friction).
Solution Analysis
Let the mass of the box be \(m\) and the total mass of the four wheels be \(m\). The system rolls down a slope of inclination \(\theta = 30^\circ\).
Case 1: Empty Cart
Total Mass: \(M_1 = m \text{ (box)} + m \text{ (wheels)} = 2m\).
The equation of motion along the slope is:
$$ 2mg \sin\theta – 4f = 2m a_1 \quad \dots(1) $$
For the wheels (rolling without slipping), Torque \(\tau = I \alpha\):
$$ f R = I_{cm} \left( \frac{a_1}{R} \right) \implies f = \frac{I_{cm}}{R^2} a_1 $$
Substituting \(f\) into eq (1):
$$ 2mg \sin\theta = \left( 2m + \frac{4I_{cm}}{R^2} \right) a_1 $$
From the initial data, we found that \(a_1 = g/3\). Substituting \(\sin 30^\circ = 0.5\):
$$ 2mg(0.5) = mg = \left( 2m + \frac{4I_{cm}}{R^2} \right) \frac{g}{3} $$
$$ 3m = 2m + \frac{4I_{cm}}{R^2} \implies \frac{4I_{cm}}{R^2} = m $$
Case 2: Loaded Cart
A load of mass \(m\) is added.
Total Mass: \(M_2 = 2m \text{ (empty)} + m \text{ (load)} = 3m\).
The new equation of motion is:
$$ 3mg \sin\theta – 4f’ = 3m a_2 $$
Using the rotational inertia term \(\frac{4I_{cm}}{R^2} = m\) derived in Case 1:
$$ 3mg \sin\theta = \left( 3m + m \right) a_2 = 4m a_2 $$
$$ 3mg(0.5) = 4m a_2 \implies 1.5g = 4 a_2 $$
$$ a_2 = \frac{1.5}{4}g = \frac{3}{8}g $$
Final Calculation
For distance \(s\), time is given by \( t = \sqrt{\frac{2s}{a}} \). The ratio of times is:
$$ \frac{t_2}{t_1} = \sqrt{\frac{a_1}{a_2}} = \sqrt{\frac{g/3}{3g/8}} = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3} $$
Given \(t_1 = 3.0\) s: