Two wheels, each of mass \(m\) and radius \(r_{w}\), are connected by an elastic coupler of effective stiffness \(k\) and undeformed length \(L\). The system rolls without slip around a circular track of radius \(R\), as shown. Derive the equations of motion of the wheel system.

Short Answer

Expert verified
In the given problem, two wheels are connected by an elastic coupler and roll on a circular track. We have found the equations of motion for this system by following these steps: 1. Define the forces and position coordinates for each wheel. 2. Apply Newton's Second Law to derive equations for horizontal and vertical forces acting on each wheel. 3. Incorporate kinematic properties of rolling motion in the equations. 4. Apply Hooke's Law to the spring forces between the wheels. 5. Substitute the values of forces and coordinates into the derived equations to find the equations of motion for the entire system. The final equations of motion are given by: $$ k(x_2 - x_1 - L) - N_1\sin\theta = m\ddot{x}_1 $$ $$ -N_2\sin\theta + k(x_1 - x_2 - L) = m\ddot{x}_2 $$ $$ N_1\sin\theta = mg\tan\theta $$ $$ N_2\sin\theta = mg\tan\theta $$ $$ \frac{dx_1}{dt} = r_w\frac{d\theta}{dt} $$ $$ \frac{dx_2}{dt} = r_w\left(\frac{d\theta}{dt} + \frac{d\alpha}{dt}\right) $$

Step by step solution

01

Define the Forces and Position Coordinates

We will start by defining the position coordinates for each wheel and the forces acting upon them. Let \(x_1\) and \(x_2\) be the horizontal coordinates of the centers of the two wheels. Let \(F_{s1}\) and \(F_{s2}\) be the spring forces on the first and second wheel, respectively, and \(\theta\) be the angle the system makes with the horizontal. The forces acting on the wheels are the gravitational force (\(mg\)), the tension force from the spring, and the normal force (\(N_1\) and \(N_2\)) exerted by the circular track on the two wheels.
02

Use Newton's Second Law

To derive the equations of motion, we will apply Newton's second law on each wheel. Newton's second law states that the net force on an object is equal to the mass of the object multiplied by its acceleration. Mathematically, it can be written as: $$ F = ma $$ Applying Newton's second law in the horizontal direction for wheel 1, we get: $$ F_{s1} - N_1\sin\theta = m\ddot{x}_1 $$ And for wheel 2: $$ -N_2\sin\theta - F_{s2} = m\ddot{x}_2 $$ Applying Newton's second law in the vertical direction for both wheels, we get: $$ N_1\cos\theta = mg $$ $$ N_2\cos\theta = mg $$
03

Incorporate Kinematics of Rolling Motion

We will now use the no-slip rolling condition, which states that the point of contact between the wheel and the circular track is at rest. For wheel 1, we have: $$ \frac{dx_1}{dt} = r_w\frac{d\theta}{dt} $$ For wheel 2, we have: $$ \frac{dx_2}{dt} = r_w\left(\frac{d\theta}{dt} + \frac{d\alpha}{dt}\right) $$ Where \(\alpha\) is the angle between the spring and the horizontal.
04

Apply Hooke's Law

To incorporate the spring elongation, we will use Hooke's Law, which states that the force exerted by a spring is proportional to its elongation. Mathematically, it is written as: $$ F_s = k\Delta x $$ The elongation of the spring is given by \(x_2 - x_1 - L\). Thus, the spring force \(F_{s1}\) and \(F_{s2}\) can be written as: $$ F_{s1} = k(x_2 - x_1 - L) $$ $$ F_{s2} = -k(x_2 - x_1 - L) $$
05

Derive the Equations of Motion

Using all the above results, we can now derive the equations of motion for the wheel system. Substituting \(F_{s1}\) and \(F_{s2}\) into the previously derived Newton's second law equations, we get the system equations: $$ k(x_2 - x_1 - L) - N_1\sin\theta = m\ddot{x}_1 $$ $$ -N_2\sin\theta + k(x_1 - x_2 - L) = m\ddot{x}_2 $$ Also, substituting the equations for \(N_1\cos\theta\) and \(N_2\cos\theta\), we get: $$ N_1\sin\theta = mg\tan\theta $$ $$ N_2\sin\theta = mg\tan\theta $$ These equations, along with the rolling motion kinematics equations for \(x_1\) and \(x_2\), constitute the equations of motion for the two-wheel system.

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