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 $\left(v_{\mathrm{L}}\right)$ relative to that of the small ball $\left(v_{\mathrm{S}}\right)$ after rolling $1\mathrm{m}?$ a) $v_{\mathrm{L}}=4v_{\mathrm{S}}$ d) $v_{\mathrm{L}}=0.5v_{\mathrm{S}}$ b) $v_{\mathrm{L}}=2v_{\mathrm{S}}$ e) $v_{\mathrm{L}}=0.25v_{\mathrm{S}}$ c) $v_{\mathrm{L}}=v_{\mathrm{S}}$

Short answer

a) v_L = v_S b) v_L = 2v_S c) v_L = 4v_S d) v_L = 0.5 v_S

Step-by-step solution

1. Find the moment of inertia of the solid sphere

For a solid sphere, the moment of inertia (I) is given by the formula: $I=\frac{2}{5}m{r}^2$, where m is the mass and r is the radius of the sphere.

2. Set up an equation for conservation of mechanical energy

As the balls roll without slipping, conservation of mechanical energy states that the total mechanical energy at the beginning (potential energy) equals the total mechanical energy at the end (kinetic energy + rotational energy). Thus, we have: $mgh=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2$ , where m: mass of the ball, g: acceleration due to gravity, h: height of the ball, v: linear velocity of the ball, I: moment of inertia, ω: angular velocity of the ball.

3. Relate linear velocity and angular velocity

The relationship between linear velocity (v) and angular velocity (ω) for a rolling sphere without slipping is given by: $v=r\omega$. This relationship will be used to eliminate ω from the conservation of mechanical energy equation.

4. Apply the conservation of mechanical energy equation to both balls and divide the equations

For the small ball, we have: $mg(1)=\frac{1}{2}m(v_S{)}^2+\frac{1}{2}(\frac{2}{5}m{r}^2)(\frac{v_S}{r}{)}^2$ And for the large ball, we have: $mg(1)=\frac{1}{2}m(v_L{)}^2+\frac{1}{2}(\frac{2}{5}m(2r{)}^2)(\frac{v_L}{2r}{)}^2$ Now, we divide the equation for the large ball by the equation for the small ball to find the ratio of their velocities, $($v_L/v_S$)$:

5. Solve for the ratio of the velocities

After dividing the equations and simplifying, we get: $\frac{v_L^2}{v_S^2}=\frac{1}{2}$ Taking the square root of both sides, we have: $v_L=\sqrt{\frac{1}{2}}{v}_S$ Comparing with the given options, we have $v_{\mathrm{L}}=0.5v_{\mathrm{S}}$. Therefore, the correct answer is (d) $v_{\mathrm{L}}=0.5v_{\mathrm{S}}$.

Key concepts

Conservation of Mechanical Energy

In physics, the conservation of mechanical energy principle is a vital concept, especially when analyzing systems in motion. It states that the total mechanical energy of a system remains constant if only conservative forces, like gravity, are acting upon it. Mechanical energy itself is the sum of potential and kinetic energy. Potential energy is related to the height of an object due to gravity, while kinetic energy pertains to its motion.

For the two rolling balls on the inclined plane, as they descend, potential energy, due to height, converts into kinetic energy, which includes both translational and rotational forms. Given no energy is lost to friction (since they roll without slipping), the initial potential energy at the top equals the kinetic energy at the bottom. Thus, we set up the equation:

• Initial potential energy: $mgh$
• Kinetic energy at the bottom: $\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2$

"$mgh=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2$" captures this sharing of energies, showcasing the additive nature of translational and rotational kinetic energy in a system. Understanding this balance is key to solving motion-related problems.

Moment of Inertia

The moment of inertia is often termed the "rotational analog of mass." It essentially measures an object's resistance to changes in its rotational motion. For a given object, it depends both on the mass and how that mass is distributed relative to the axis of rotation.

In the case of the spheres rolling down the incline, their moment of inertia is crucial in determining their rotational kinetic energy. For a solid sphere, the moment of inertia $I$ is calculated with the formula: $I=\frac{2}{5}m{r}^2$. This represents how the mass ($m$) and radius ($r$) influence their ability to rotate. The larger the mass or radius, the greater the moment of inertia, thus making it more challenging to spin the object.

When the large and small balls roll, their motions are inherently tied to their inertia. Given the large ball has a greater radius, its moment of inertia (being $\frac{2}{5}m(2r{)}^2$) is significantly greater than the small ball's inertia. This influences how fast it can roll, emphasizing the importance of moment of inertia in analyzing and predicting the dynamics of rolling objects.

Rolling Motion

Rolling motion is a fascinating mechanical concept where an object moves along a surface without slipping. This means that every point of the object in contact with the surface at any given time is momentarily stationary. It incorporates both translational and rotational motion, often leading to efficient energy conservation.

For rolling objects like our spheres, the linear velocity $v$ and angular velocity $\omega$ are linked by the relationship $v=r\omega$. This relationship ensures that for every complete rotation, the object moves forward a distance equal to its circumference. Without slipping, this means normalizing the object's motion to both rotate and translate in perfect harmony.

• Linear motion: Body moves in a straight path.
• Rotational motion: Body spins around its axis.

The consistent application of these principles helps determine the path and speed of rolling objects, such as calculating the velocity ratio between the two different spheres on the incline. Understanding rolling motion is essential for deciphering how objects move seamlessly with efficiency, much like our steel balls descending without loss of energy to friction.