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 their accelerations is \(\frac{a_{\text {ring }}}{a_{\text {sphere }}} = \frac{7}{5}\).

Step-by-step solution

Determine moment of inertia for each object

The moment of inertia for a solid sphere is given by \(I_{\text{sphere}} = \frac{2}{5}mr^2\), and for a thin ring, it is given by \(I_{\text{ring}} = mr^2\). We will use these values to find the kinetic energy of each object.

Set up equations for total energy of each object

The total energy of the sphere and the ring are given as follows: \(sphere: \, \frac{1}{2}m_{s}v_{s}^2 + \frac{1}{2}I_{\text{sphere}} \omega_{s}^2 = m_{s}gh\) \(ring: \, \frac{1}{2}m_{r}v_{r}^2 + \frac{1}{2}I_{\text{ring}} \omega_{r}^2 = m_{r}gh\) These equations include the conservation of mechanical energy, where the potential energy at the top of the ramp is being converted into linear kinetic energy and rotational kinetic energy.

Express velocities in terms of angular velocities

Since both objects are rolling without slipping, we can relate linear velocities to angular velocities using the relationship \(v = r\omega\). Therefore, we can rewrite the energies as: \(sphere: \, \frac{1}{2}m_{s}(r\omega_{s})^2 + \frac{1}{2}\frac{2}{5}m_sr^2\omega_{s}^2 = m_{s}gh\) \(ring: \, \frac{1}{2}m_{r}(r\omega_{r})^2 + \frac{1}{2}m_{r}r^2\omega_{r}^2 = m_{r}gh\)

Solve the energy equations for angular velocities

Now, we solve for \(\omega_{s}\) and \(\omega_{r}\): \(sphere: \, \omega_{s}^2 = \frac{10gh}{7r^2}\) \(ring: \, \omega_{r}^2 = \frac{2gh}{r^2}\)

Calculate accelerations

Using the relationships \(a_{s} = r\alpha_{s}\) and \(a_{r} = r\alpha_{r}\), with \(\alpha\) being the angular acceleration, and remembering that \(a = r\alpha\): \(solid\, sphere: \, a_{\text{sphere}} = r \frac{d\omega_{s}}{dt} = \frac{10gh}{7} \) \(thin\, ring: \, a_{\text{ring}} = r \frac{d\omega_{r}}{dt} = 2gh\)

Calculate the ratio of accelerations

Now, we can find the desired ratio of accelerations: \(\frac{a_{\text{ring}}}{a_{\text{sphere}}} = \frac{2gh}{\frac{10gh}{7}} = \frac{14}{10} = \frac{7}{5}\) The ratio of their accelerations is \(\frac{a_{\text {ring }}}{a_{\text {sphere }}} = \frac{7}{5}\).

Key concepts

Moment of Inertia

The moment of inertia essentially works as inertia does for linear motion, but for rotational motion. It's a way to quantify how much torque is needed for a particular angular acceleration on a rigid body. It all comes down to how the mass of an object is distributed relative to the axis of rotation; the further the mass is spread from the axis, the larger the moment of inertia and the harder it is to get it rotating. For example, in our exercise, a solid sphere has a moment of inertia given by \(I_{\text{sphere}} = \frac{2}{5}mr^2\), and a thin ring, with all its mass at the edge, has a larger moment of inertia, \(I_{\text{ring}} = mr^2\). This difference in distribution is why their accelerations would differ when they roll down the ramp.

Conservation of Mechanical Energy

Conservation of mechanical energy is a principle stating that if in a closed system with no non-conservative forces (like friction) acting on it, the total mechanical energy (kinetic plus potential energy) remains constant. In the context of our exercise, as the solid sphere and the thin ring roll down the ramp, their potential energy is converted into two forms of kinetic energy — linear and rotational — but the total amount of mechanical energy remains the same. This conservation allows us to set up the energy equations for both the sphere and the ring and solve for their variables, as energy at the top of the ramp (potential energy) is equal to energy at any subsequent moment during the roll (sum of linear and rotational kinetic energies).

Rolling Without Slipping

Rolling without slipping is a condition where the object rolls on a surface in such a way that there is no relative motion between the point of contact on the rolling object and the surface it's rolling on. This means that the static frictional force is sufficient to prevent slipping. Rolling without slipping allows us to connect linear motion and rotational motion through the relationship \(v = r\omega\), where \(v\) is linear velocity, \(r\) is the radius of the rolling object, and \(\omega\) is its angular velocity. This relationship is crucial in solving problems involving rolling objects because it links linear and angular acceleration, which is used to solve for the accelerations as shown in the exercise.