Question

In another race, a solid sphere and a thin ring roll without slipping from rest down a ramp that makes angle \(\theta\) with the horizontal. Find the ratio of their accelerations, \(a_{\text {ring }} / a_{\text {sphere }}\)

Short answer

Answer: The ratio of accelerations of a solid sphere and a thin ring rolling without slipping down a ramp at angle θ with the horizontal is 7/10.

Step-by-step solution

1. Write down the expressions of the forces acting on the objects

The gravitational force \(F_g\) acts vertically downwards on both the sphere and the ring. It can be decomposed into two components; one along the incline (\(F_\parallel\) which causes the object to roll) and one perpendicular to the incline (\(F_\perp\) which doesn't contribute to the motion). We will find \(F_\parallel\) for both objects. For both objects, we have: \(F_g = mg\) where \(m\) is the mass of the object (sphere or ring) and \(g\) is the acceleration due to gravity. The component of gravitational force along the incline, in terms of \(\theta\) is given by: \(F_\parallel = mg\sin\theta\)

2. Set up equations of motion

For rolling without slipping, we need to equate the translational and rotational motion. We use Newton's second law for translational motion and the torque equation for rotational motion: For translational motion, we have: $$F_\parallel = ma$$ For rotational motion, we have: $$I\alpha=R \cdot F_\parallel$$ where \(I\) is the moment of inertia of the object, \(\alpha\) is the angular acceleration, and \(R\) is the radius. Since the objects are rolling without slipping, we can relate the linear and angular acceleration using the formula: $$a=R\alpha$$

3. Calculate the moment of inertia for each object

Now, we need to find the moment of inertia, \(I\), for each object. The moment of inertia depends on the distribution of mass in the object. For a solid sphere of mass \(m\) and radius \(R\), its moment of inertia about an axis through the center is given by: \(I_{\text {sphere }}=\frac{2}{5}mR^2\) For a thin ring of mass \(m\) and radius \(R\), its moment of inertia about an axis perpendicular to the plane of the ring and passing through the center is given by: \(I_{\text {ring }}=mR^2\)

4. Solve the equations for the accelerometer and find the ratio

Now we need to solve the equations of motion for both objects to find the accelerations for each object: For the sphere: $$\begin{aligned} m a_{\text {sphere }} &=m g \sin \theta \\ \frac{2}{5} m R^2 \alpha_{\text {sphere }} &=R \cdot a_{\text {sphere }} \\ a_{\text{sphere}} &= \frac{5}{7}g\sin\theta \end{aligned}$$ For the ring: $$\begin{aligned} m a_{\text {ring }} &=m g \sin \theta \\ m R^2 \alpha_{\text {ring }} &=R \cdot a_{\text {ring }} \\ a_{\text{ring}} &= \frac{1}{2}g\sin\theta \end{aligned}$$ Now we can find the desired ratio: $$\frac{a_{\text {ring }} }{a_{\text {sphere }}}=\frac{\frac{1}{2} g \sin \theta}{\frac{5}{7} g \sin \theta}=\boxed{\frac{7}{10}}$$