A small sphere rolls down without slipping from the top of a track in a vertical plane. The track ha

Q: Why do we include rotational kinetic energy?

Because the sphere is rolling without slipping, it has both translational motion (moving forward) and rotational motion (spinning). Both require energy, which comes from the loss in gravitational potential energy.

Q: Why is the work done by friction zero during pure rolling?

In pure rolling, the point of contact between the sphere and the track is instantaneously at rest relative to the track. Since there is no relative displacement at the point of application of the friction force, the work done by static friction is zero.

Q: Does the sphere's rotation affect its projectile trajectory?

No. The translational motion of the center of mass is governed only by gravity (ignoring air resistance). The rotation does not create any additional aerodynamic forces (like the Magnus effect) in this idealized physics problem.

Q: Why does the sphere continue to rotate in the air?

Once in the air, the only force acting on the sphere is gravity. Gravity acts exactly through the center of mass, so it creates zero torque about the center of mass. With no net torque, angular momentum is conserved, and the sphere keeps spinning at a constant rate.

Q: What if the sphere was sliding instead of rolling?

If it was sliding without friction, all the potential energy would convert into translational kinetic energy ($mgh = \frac{1}{2}mv^2$), resulting in a higher launch velocity and a longer horizontal range. It would not rotate.

Question

A small sphere rolls down without slipping from the top of a track in a vertical plane. The track has an elevated section and a horizontal part. The horizontal part is $1.0	ext{ m}$ above the ground level and the top of the track is $2.6	ext{ m}$ above the ground. Find the distance on the ground with respect to the point $B$ (which is vertically below the end of the track as shown in figure) where the sphere lands. During its flight as a projectile, does the sphere continue to rotate about its centre of mass? Explain.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing at the top of a roller coaster track, holding a small solid sphere. The track starts at a towering height of $2.6	ext{ m}$ above the ground. It swoops down and levels out at a height of $1.0	ext{ m}$ before abruptly ending, launching anything on it into the air.

When you release the sphere, it doesn't just slide down; it rolls. This is a crucial detail. Rolling without slipping means that the point of the sphere in contact with the track is momentarily at rest. Because there is no relative motion at the contact point, the work done by friction is exactly zero.

This beautiful fact allows us to use the principle of conservation of mechanical energy. The sphere starts with a certain amount of gravitational potential energy. As it descends, this energy is transformed, but not lost. The effective height it drops before launching is the difference between the starting height and the launch height:

$$h = 2.6	ext{ m} - 1.0	ext{ m} = 1.6	ext{ m}$$

### The Master Equation: Energy Conservation

As the sphere rolls down, its potential energy converts into kinetic energy. But because it's rolling, this kinetic energy comes in two flavors: translational (moving forward) and rotational (spinning around its center).

For a solid sphere, the moment of inertia is $I = \frac{2}{5}mr^2$. The relationship between rotational kinetic energy ($K_R$) and translational kinetic energy ($K_T$) is elegantly simple:

$$K_R = \frac{1}{2}I\omega^2 = \frac{1}{2}\left(\frac{2}{5}mr^2ight)\left(\frac{v}{r}ight)^2 = \frac{2}{5}\left(\frac{1}{2}mv^2ight) = \frac{2}{5}K_T$$

This means the rotational energy is always $40\%$ of the translational energy. The total kinetic energy is the sum of both:

$$K_{total} = K_T + K_R = K_T + \frac{2}{5}K_T = \frac{7}{5}K_T$$

By conservation of energy, the loss in potential energy equals the gain in total kinetic energy:

$$mgh = \frac{7}{5}K_T$$

Substituting $K_T = \frac{1}{2}mv^2$, we can solve for the velocity $v$ at the moment the sphere leaves the track:

$$\frac{1}{2}mv^2 = \frac{5}{7}mgh$$

$$v = \sqrt{\frac{10}{7}gh}$$

Plugging in the values $g = 9.8	ext{ m/s}^2$ and $h = 1.6	ext{ m}$:

$$v = \sqrt{\frac{10}{7} 	imes 9.8 	imes 1.6} \approx 4.73	ext{ m/s}$$

### Taking Flight: Projectile Motion

The moment the sphere leaves the track, it becomes a projectile. It shoots off horizontally with a velocity of $4.73	ext{ m/s}$ from a height of $H = 1.0	ext{ m}$.

To find out how far it travels horizontally, we first need to know how long it stays in the air. The vertical motion is governed entirely by gravity, starting from rest in the vertical direction. We use the classic kinematic equation:

$$H = \frac{1}{2}gt^2 \implies t = \sqrt{\frac{2H}{g}}$$

Substituting $H = 1.0	ext{ m}$:

$$t = \sqrt{\frac{2 	imes 1.0}{9.8}} \approx 0.45	ext{ s}$$

During this $0.45$ seconds, the sphere continues to move horizontally at a constant speed of $4.73	ext{ m/s}$. The horizontal distance (range) $BC$ is simply velocity multiplied by time:

$$BC = v 	imes t = 4.73 	imes 0.45 \approx 2.13	ext{ m}$$

### The Spin Continues: Conservation of Angular Momentum

Now for the conceptual twist: does the sphere keep spinning while it's flying through the air?

To answer this, we must look at the forces acting on the sphere during its flight. The only force is gravity (ignoring air resistance). Crucially, gravity acts exactly through the center of mass of the sphere.

Torque is defined as force multiplied by the perpendicular distance from the axis of rotation. Since the force of gravity passes directly through the center of mass, the perpendicular distance is zero. Therefore, the torque about the center of mass is zero:

$$	au_{cm} = 0$$

According to Newton's second law for rotation, if the net external torque is zero, the angular momentum must remain constant.

$$\vec{L}_{cm} = 	ext{constant}$$

Because the moment of inertia of the sphere doesn't change, its angular velocity $\omega$ must also remain constant. The sphere will continue to spin in the air at the exact same rate it was spinning when it left the track! It's a beautiful demonstration of how translational and rotational motions become completely independent once the sphere is airborne.

Visualized Solution

Analyzing the Setup

Conservation of Mechanical Energy

Kinetic Energy of a Rolling Sphere

Translational Velocity at the Edge

Calculating the Velocity

Projectile Motion: Time of Flight

Horizontal Range

Rotation During Flight

The Sigma Insight: Work and Energy in Rotational Motion

Analyzing the Setup

The Master Equation

Taking Flight

The Spin Continues

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