A solid sphere and solid cylinder of identical radii approach an incline with the same linear veloci

Q: Why do we consider both translational and rotational kinetic energy?

Because the body is rolling without slipping, it possesses both a linear velocity (moving forward) and an angular velocity (spinning). Both motions require energy, so the total kinetic energy is the sum of both.

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

It means the point of contact with the ground is instantaneously at rest. Mathematically, this gives us the constraint equation $v = \omega R$, linking the linear velocity of the center of mass to the angular velocity.

Q: Why doesn't the mass or radius affect the final ratio?

If you look closely at the equations, both mass $m$ and radius $R$ cancel out when we substitute the moment of inertia and equate kinetic energy to potential energy. The height depends only on the shape (mass distribution) and the initial velocity.

Q: Would a hollow sphere climb higher or lower than a solid sphere?

A hollow sphere has a larger moment of inertia ($\frac{2}{3}mR^2$) compared to a solid sphere ($\frac{2}{5}mR^2$). This means it stores more rotational kinetic energy for the same linear speed, so it would climb higher!

Q: Is friction doing any work here?

In pure rolling, the point of contact is instantaneously at rest, meaning there is no relative displacement between the surfaces at the point of contact. Therefore, the work done by static friction is zero, and mechanical energy is conserved.

Question

A solid sphere and solid cylinder of identical radii approach an incline with the same linear velocity (see figure). Both roll without slipping all throughout. The two climb maximum heights $h_{	ext{sph}}$ and $h_{	ext{cyl}}$ on the incline. The ratio $\frac{h_{	ext{sph}}}{h_{	ext{cyl}}}$ is given by

Options:

(A) $\frac{2}{5}$

(B) $\frac{14}{15}$

(C) $1$

(D) $\frac{4}{5}$

Sigma Percentile · Accepted Answer

### The Rolling Challenge

Imagine you are standing at the base of a ramp, holding a solid sphere in one hand and a solid cylinder in the other. Both have the exact same radius. You roll them towards the ramp with the exact same linear velocity, $v$. The question is: which one will climb higher, and what is the exact ratio of their maximum heights?

This is a classic problem of rolling motion, where objects don't just slide—they spin! Because they are rolling without slipping, we don't have to worry about energy lost to friction. The total mechanical energy of the system is perfectly conserved.

### Unpacking the Kinetic Energy

When a body rolls, its total kinetic energy is the sum of its translational kinetic energy (moving forward) and its rotational kinetic energy (spinning around its center).

$$K_{	ext{total}} = K_{	ext{trans}} + K_{	ext{rot}} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$$

For pure rolling, the linear velocity $v$ and the angular velocity $\omega$ are locked together by the relation $v = \omega R$. Substituting this into our energy equation gives us a beautiful, unified expression:

$$K_{	ext{total}} = \frac{1}{2}mv^2 + \frac{1}{2}I\left(\frac{v}{R}ight)^2$$

### Sphere vs. Cylinder

Now, let's look at our two competitors. The solid sphere has a moment of inertia $I_{	ext{sph}} = \frac{2}{5}mR^2$. Plugging this into our total energy equation:

$$K_{	ext{sph}} = \frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{2}{5}mR^2ight)\left(\frac{v}{R}ight)^2 = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2$$

Next, the solid cylinder. Its mass is distributed differently, giving it a moment of inertia $I_{	ext{cyl}} = \frac{1}{2}mR^2$. Let's find its total kinetic energy:

$$K_{	ext{cyl}} = \frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{1}{2}mR^2ight)\left(\frac{v}{R}ight)^2 = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2$$

Notice how the cylinder has a slightly higher total kinetic energy ($\frac{3}{4} = 0.75$) compared to the sphere ($\frac{7}{10} = 0.70$) for the same linear velocity! This means the cylinder has more energy stored in its rotation.

### The Final Showdown

As they climb the ramp, all this kinetic energy is converted into gravitational potential energy, $mgh$.

For the sphere:
$$mgh_{	ext{sph}} = \frac{7}{10}mv^2 \implies h_{	ext{sph}} = \frac{7v^2}{10g}$$

For the cylinder:
$$mgh_{	ext{cyl}} = \frac{3}{4}mv^2 \implies h_{	ext{cyl}} = \frac{3v^2}{4g}$$

To find the ratio of their maximum heights, we simply divide the two expressions. The $v^2$ and $g$ terms cancel out perfectly:

$$\frac{h_{	ext{sph}}}{h_{	ext{cyl}}} = \frac{\frac{7v^2}{10g}}{\frac{3v^2}{4g}} = \frac{7}{10} 	imes \frac{4}{3} = \frac{28}{30} = \frac{14}{15}$$

And there we have it! The ratio is $\frac{14}{15}$. The cylinder climbs slightly higher because its mass distribution gives it a larger moment of inertia, allowing it to store more rotational kinetic energy for the same linear speed.

Select Answer:

Visualized Solution

Visualizing the Setup

Conservation of Energy

Total Kinetic Energy

Kinetic Energy of Solid Sphere

Kinetic Energy of Solid Cylinder

Calculating Maximum Heights

The Final Ratio

The Radius of Gyration Shortcut

The Sigma Insight: Rolling Motion

The Rolling Challenge

Unpacking the Kinetic Energy

Sphere vs

The Final Showdown

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