Question

Two solid steel balls, one small and one large, are on an inclined plane. The large ball has a diameter twice as large as that of the small ball. Starting from rest, the two balls roll without slipping down the incline until their centers of mass are \(1 \mathrm{~m}\) below their starting positions. What is the speed of the large ball \(\left(v_{\mathrm{I}}\right)\) relative to that of the small ball \(\left(v_{\mathrm{c}}\right)\) after rolling \(1 \mathrm{~m} ?\) a) \(v_{\mathrm{L}}=4 v_{\mathrm{S}}\) b) \(v_{\mathrm{L}}=2 v_{\mathrm{S}}\) c) \(v_{\mathrm{L}}=v_{\mathrm{S}}\) d) \(v_{\mathrm{L}}=0.5 v_{\mathrm{s}}\) e) \(v_{\mathrm{L}}=0.25 v_{\mathrm{S}}\)

Short answer

Based on this step-by-step solution, the short answer would be: The relative speed of the large ball with respect to the small ball after they have rolled down 1 meter is \(\sqrt{2}\) times greater than the small ball's speed, approximately 1.41 times faster.

Step-by-step solution

Identify the relevant equations

We will use the equations for conservation of mechanical energy and moment of inertia for solid spheres. The conservation of mechanical energy equation is: \(K_i + U_i = K_f + U_f\) where \(K\) represents kinetic energy and \(U\) represents potential energy. The moment of inertia for a solid sphere is: \(I = \frac{2}{5} m R^2\) where \(I\) is the moment of inertia, \(m\) is the mass, and \(R\) is the radius of the sphere.

Apply conservation of mechanical energy to both balls

We will apply the conservation of mechanical energy equation to both balls separately. For the small ball: \(\frac{1}{2} m_sv_s^2 + \frac{1}{2} I_s \omega_s^2 = m_sgh\) For the large ball: \(\frac{1}{2} m_lv_l^2 + \frac{1}{2} I_l \omega_l^2 = m_lgh\)

Calculate the moment of inertia and angular velocities of both balls

To find the angular velocities for both balls, we apply the following equation: \(v = R \omega\) For the small ball: \(\omega_s = \frac{v_s}{R_s}\) For the large ball: \(\omega_l = \frac{v_l}{R_l}\) Now, calculate the moment of inertia for both balls. For the small ball: \(I_s = \frac{2}{5} m_s R_s^2\) For the large ball: \(I_l = \frac{2}{5} m_l(\frac{1}{2}R_s)^2\) Since the large ball has a diameter twice as large as the small ball, its mass is 8 times larger, so \(m_l = 8m_s\).

Substitute the moment of inertia and angular velocities into the energy equations

Now, substitute the moment of inertia and angular velocities back into the conservation of mechanical energy equations. For the small ball: \(\frac{1}{2} m_sv_s^2 + \frac{1}{2}\left(\frac{2}{5} m_s R_s^2\right)\frac{v_s^2}{R_s^2} = m_sgh\) For the large ball: \(\frac{1}{2} m_lv_l^2 + \frac{1}{2}\left(\frac{2}{5} (8m_s)\frac{1}{4} R_s^2\right)\frac{v_l^2}{\left(\frac{1}{2}R_s\right)^2} = 8m_sgh\)

Solve the equations

Solve the equations for \(v_s\) and \(v_l\). For the small ball: \(\frac{7}{10}m_sv_s^2 = m_sgh\) \(v_s^2 = \frac{10}{7}gh\) For the large ball: \(\frac{7}{5}m_lv_l^2 = 8m_sgh\) \(v_l^2 = \frac{20}{7}gh\) Now, let's find the ratio of \(v_l\) to \(v_s\): \(v_l = \sqrt{\frac{20}{7}gh}\) \(v_s = \sqrt{\frac{10}{7}gh}\) \(\frac{v_l}{v_s} = \frac{\sqrt{\frac{20}{7}gh}}{\sqrt{\frac{10}{7}gh}} = \sqrt{2}\) This means that the speed of the large ball is \(\sqrt{2}\) times greater than the speed of the small ball, which is closest to answer choice (d) \(v_{\mathrm{L}}=0.5 v_{\mathrm{s}}\).

Key concepts

Conservation of Mechanical Energy

The conservation of mechanical energy is a fundamental principle in physics, stating that the total mechanical energy in a closed system remains constant if only conservative forces are acting. Mechanical energy itself is the sum of kinetic energy, which is energy due to motion, and potential energy, which is energy due to position or configuration.

In the context of our exercise involving rolling balls on an inclined plane, the mechanical energy of each ball is conserved as they roll downward—this is because the force of gravity is a conservative force. As the balls move from higher to lower positions, their potential energy is converted into kinetic energy. However, there is a key point to remember; in this case, the kinetic energy is split into translational kinetic energy (\frac{1}{2}mv^2) and rotational kinetic energy (\frac{1}{2}I\theta^2). The conservation of mechanical energy equation is written as:

\[\begin{equation} K_i + U_i = K_f + U_f \text{where}\begin{align*}K_i\&\text{is the initial kinetic energy,}\U_i\&\text{is the initial potential energy,}\K_f\&\text{is the final kinetic energy, and}\U_f\&\text{is the final potential energy.}\text{For rolling objects, it's crucial to include both forms of kinetic energy in these calculations.} By applying this principle, we were able to solve for the final velocities of the balls in our exercise and determine how their speeds compared.

Moment of Inertia

The moment of inertia, symbolized by 'I', is a measure of an object's resistance to changes in its rotational motion and, more specifically, how this resistance is distributed throughout the object. It plays a similar role in rotational motion as mass does in linear motion. For rigid bodies, the moment of inertia depends not only on the mass but also on how that mass is distributed in relation to the axis of rotation.

Different shapes have different moments of inertia formulas. For a solid sphere, as addressed in our exercise, the moment of inertia is given by:
\[\begin{equation} I = \frac{2}{5}mR^2 \text{where}\begin{align*}m\&\text{is the mass of the sphere and}\R\&\text{is the radius of the sphere.}\text{The factor of}\frac{2}{5}\text{is a result of the spherical mass distribution.}\end{align*}\end{equation}\]
For our steel balls rolling without slipping, different masses and sizes meant that we needed to account for the different moments of inertia when calculating the mechanical energy present in their rotational motion.

Rolling Without Slipping

Rolling without slipping is a term used in rotational motion to describe a scenario where an object rolls on a surface without any sliding. When an object is rolling without slipping, there is a static friction force present between the object and the surface which prevents sliding. This static friction does not do work, as the point of contact is momentarily at rest as it touches the surface.

This concept is crucial for solving problems involving rolling motion because it defines a relationship between the translational velocity (v) of the object's center of mass and its angular velocity (\theta). Specifically, the relationship is given by: \[\begin{equation} v = R\theta \text{where}\begin{align*}v\&\text{is the linear velocity,}\R\&\text{is the radius of the object, and}\theta\&\text{is the angular velocity.}\end{align*}\end{equation}\]
In the exercise, ensuring that the balls roll without slipping allows us to use this relationship to connect linear and angular quantities, ultimately helping us to understand the energy transformation during the balls' descent.