A sphere of radius a and mass m rolls along a horizontal plane with constant speed v0. It encounters

Q: Why do we include rotational kinetic energy?

Because the sphere is rolling without slipping, it possesses both translational kinetic energy (due to its forward motion) and rotational kinetic energy (due to its spinning). Both contribute to the total energy that allows it to climb the incline.

Q: What does 'rolling without slipping' mean mathematically?

It means the velocity of the center of mass $v_0$ and the angular velocity $\omega$ are perfectly synced by the relation $v_0 = \omega a$, where $a$ is the radius of the sphere.

Q: Why is the work done by friction zero?

In pure rolling, the point of contact between the sphere and the surface is instantaneously at rest. Since there is no relative displacement at the point of application of the friction force, it does zero work.

Q: What if the incline was frictionless?

If the incline was perfectly smooth, the friction force would be zero. The sphere would continue to spin at its initial angular velocity $\omega$ even as it climbs, meaning only its translational kinetic energy would convert to potential energy.

Q: Why doesn't the mass or radius of the sphere affect the final answer?

Both the kinetic energy and the potential energy are directly proportional to the mass $m$. Similarly, the radius $a$ cancels out when we substitute $\omega = v_0/a$ into the rotational kinetic energy term. Thus, the final distance depends only on the initial speed, gravity, and the incline angle.

Question

A sphere of radius $a$ and mass $m$ rolls along a horizontal plane with constant speed $v_0$. It encounters an inclined plane at angle $	heta$ and climbs upwards. Assuming that it rolls without slipping, how far up the sphere will travel?

Options:

(A) $\frac{v_0^2}{2g \sin	heta}$

(B) $\frac{v_0^2}{5g \sin	heta}$

(C) $\frac{10v_0^2}{7g \sin	heta}$

(D) $\frac{2v_0^2}{5g \sin	heta}$

Sigma Percentile · Accepted Answer

### The Setup: A Rolling Sphere

Imagine a solid sphere of mass $m$ and radius $a$ rolling smoothly across a flat horizontal floor. It is moving with a constant linear speed $v_0$.

Because it is rolling without slipping, it isn't just gliding forward; it is also spinning around its center. This means it possesses both translational and rotational kinetic energy.

Suddenly, the sphere encounters a ramp inclined at an angle $	heta$. Our goal is to find out exactly how far up this ramp the sphere will travel before momentarily coming to a halt.

### The Master Equation: Conservation of Energy

Since the sphere rolls without slipping, the point of contact with the ground is instantaneously at rest. This means friction does no work, and mechanical energy is perfectly conserved.

We can equate the total mechanical energy at the bottom of the incline to the total mechanical energy at the maximum height.

At the bottom, the energy is entirely kinetic. At the highest point, the sphere stops, meaning all that kinetic energy has been converted into gravitational potential energy.

$$E_{	ext{initial}} = E_{	ext{final}}$$

### Calculating the Initial Energy

Let's break down the initial kinetic energy. It is the sum of the translational kinetic energy and the rotational kinetic energy.

$$E_{	ext{initial}} = \frac{1}{2}mv_0^2 + \frac{1}{2}I\omega^2$$

For a solid sphere, the moment of inertia $I$ about its center is $\frac{2}{5}ma^2$. Because it rolls without slipping, its angular velocity $\omega$ is tightly coupled to its linear speed by the relation $\omega = \frac{v_0}{a}$.

Substituting these values into our energy equation gives us the total initial energy.

$$E_{	ext{initial}} = \frac{1}{2}mv_0^2 + \frac{1}{2}\left(\frac{2}{5}ma^2ight)\left(\frac{v_0}{a}ight)^2$$

Notice how the radius $a$ beautifully cancels out! This leaves us with the rotational kinetic energy as $\frac{1}{5}mv_0^2$. Adding this to the translational part, we get the total initial energy.

$$E_{	ext{initial}} = \frac{1}{2}mv_0^2 + \frac{1}{5}mv_0^2 = \frac{7}{10}mv_0^2$$

### The Climb: Reaching Maximum Height

Now, let's look at the final state. At the maximum height $h$, the sphere has stopped moving and spinning. All its energy is now stored as gravitational potential energy.

$$E_{	ext{final}} = mgh$$

By equating the initial and final energies, we can solve for the maximum vertical height $h$ reached by the sphere.

$$\frac{7}{10}mv_0^2 = mgh$$

The mass $m$ cancels out from both sides, showing that the height is independent of the sphere's mass. Solving for $h$, we get:

$$h = \frac{7v_0^2}{10g}$$

### The Final Distance

The question asks for the distance $l$ traveled *along* the incline, not just the vertical height. We can relate $h$ and $l$ using simple trigonometry.

From the right-angled triangle formed by the incline, we know that $\sin	heta = \frac{h}{l}$. Rearranging this gives us $l = \frac{h}{\sin	heta}$.

Substituting our expression for $h$ into this geometric relation yields the final answer.

$$l = \frac{7v_0^2}{10g \sin	heta}$$

**Notice that none of the given options in the original question match this exact expression!** The closest option is the reciprocal of the fraction, making this a classic bonus question where all options were technically incorrect.

Animated Solution for Physics - Rotational Motion: A sphere of radius $a$ and mass $m$ rolls along a horizontal plane with constant speed $v_{0}$ . It encounters an inclined plane at angle $θ$ and climbs upwards. Assuming that it rolls without slipping, how far up the sphere will travel?

Select Answer:

Visualized Solution

$v_{0}$

$E_{initial} = E_{final}$

$E_{initial}$

$I and ω$

$E_{initial} = \frac{7}{10} m v_{0}^{2}$

$E_{final} = m g h$

$h = \frac{7 v _{0}^{2}}{10 g}$

$l = \frac{h}{s i n θ}$

$l = \frac{7 v _{0}^{2}}{10 g s i n θ}$

$Smooth Incline$

The Sigma Insight: Rolling Motion

The Setup

The Master Equation

Calculating the Initial Energy

The Climb

The Final Distance

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Select Answer:

Visualized Solution

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Smooth Incline