Two bodies, a ring and a solid cylinder of same material are rolling down without slipping an inclin

Q: Why is mechanical energy conserved if there is friction on the incline?

In pure rolling (rolling without slipping), the point of contact between the body and the surface is instantaneously at rest. Because there is no relative motion at the contact point, the static friction force does zero work. Hence, mechanical energy is conserved.

Q: Why does the solid cylinder reach the bottom faster than the ring?

The solid cylinder has its mass distributed closer to its center, giving it a smaller moment of inertia compared to the ring. This means it requires less energy to rotate, allowing more of the initial potential energy to be converted into translational kinetic energy (linear speed).

Q: Does the mass or radius of the objects affect their final velocity?

Surprisingly, no! If you look at the energy equations, both mass $m$ and radius $R$ cancel out completely. The final velocity depends only on the height $h$, gravity $g$, and the shape of the object (specifically, its geometric mass distribution factor).

Q: Can I solve this using forces and kinematics instead of energy?

Absolutely. You can draw a free body diagram, write Newton's second law for translation ($mg \sin\theta - f = ma$) and rotation ($fR = I_{CM}\alpha$), and use the rolling constraint ($a = \alpha R$). This will give you the acceleration $a$, which you can plug into $v^2 = u^2 + 2as$. However, the energy method is generally much faster and less prone to algebraic errors.

Q: What is the 'Instantaneous Center of Rotation' (ICOR) trick used here?

In pure rolling, the body can be viewed as purely rotating about its point of contact with the ground at any given instant. By calculating the moment of inertia about this contact point ($I_{contact} = I_{CM} + mR^2$), you can write the total kinetic energy as a single rotational term $\frac{1}{2} I_{contact} \omega^2$, bypassing the need to add translational and rotational kinetic energies separately.

Question

Two bodies, a ring and a solid cylinder of same material are rolling down without slipping an inclined plane. The radii of the bodies are same. The ratio of velocity of the centre of mass at the bottom of the inclined plane of the ring to that of the cylinder is $\frac{\sqrt{x}}{2}$. Then, the value of $x$ is …… .

Sigma Percentile · Accepted Answer

Imagine a grand race on a frictionless inclined plane. The competitors? A perfectly uniform ring and a solid cylinder, both forged from the same material and sharing the exact same radius. They are released simultaneously from the same height, $h$. The rules of the race are simple: they must roll down without slipping. Who will cross the finish line first, and by what margin? Let's dive into the elegant physics of rolling motion to find out.

### The Physics of Pure Rolling

When a rigid body rolls down an incline without slipping, the point of contact with the surface is instantaneously at rest. This means that static friction does no work, and the total mechanical energy of the system is perfectly conserved. The gravitational potential energy at the top transforms entirely into kinetic energy at the bottom.

However, this kinetic energy is split into two forms: translational kinetic energy (moving forward) and rotational kinetic energy (spinning). A brilliant shortcut to handle this is to consider the body as purely rotating about its instantaneous point of contact. The total kinetic energy can then be written simply as:

$$KE = \frac{1}{2} I_{contact} \omega^2$$

By the parallel axis theorem, the moment of inertia about the contact point is $I_{contact} = I_{CM} + mR^2$. Furthermore, the condition for pure rolling links the linear velocity $v$ to the angular velocity $\omega$ via the relation $v = \omega R$.

### Analyzing the Ring

Let's evaluate the ring first. A ring has all its mass concentrated at its rim, giving it the maximum possible moment of inertia for a given mass and radius: $I_{CM} = mR^2$.

Applying the parallel axis theorem, its moment of inertia about the contact point becomes:

$$I_{contact} = mR^2 + mR^2 = 2mR^2$$

Equating the initial potential energy to the final kinetic energy:

$$mgh = \frac{1}{2} (2mR^2) \left(\frac{v_R}{R}\right)^2$$

Notice how beautifully the mass $m$ and radius $R$ cancel out! Solving for the velocity of the ring, $v_R$, we get:

$$v_R = \sqrt{gh}$$

### Analyzing the Solid Cylinder

Now, let's turn our attention to the solid cylinder. Unlike the ring, its mass is distributed uniformly throughout its volume, meaning more mass is closer to the axis of rotation. This results in a lower moment of inertia: $I_{CM} = \frac{1}{2}mR^2$.

Its moment of inertia about the contact point is:

$$I_{contact} = \frac{1}{2}mR^2 + mR^2 = \frac{3}{2}mR^2$$

Again, applying the conservation of energy:

$$mgh = \frac{1}{2} \left(\frac{3}{2}mR^2\right) \left(\frac{v_C}{R}\right)^2$$

$$mgh = \frac{3}{4} m v_C^2$$

Solving for the velocity of the cylinder, $v_C$, we find:

$$v_C = \sqrt{\frac{4gh}{3}}$$

Comparing the two, $\sqrt{1.33gh}$ is clearly greater than $\sqrt{1gh}$. The solid cylinder wins the race! Because it has a lower moment of inertia, it 'spends' less of its potential energy on spinning, leaving more energy available for linear translation.

### The Final Ratio

The problem asks for the ratio of the velocity of the ring to that of the cylinder:

$$\frac{v_R}{v_C} = \frac{\sqrt{gh}}{\sqrt{\frac{4gh}{3}}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$$

We are given that this ratio is equal to $\frac{\sqrt{x}}{2}$. By direct comparison, it is evident that **$x = 3$**.

Enter Numerical Value:

Visualized Solution

\text{Rolling on an Incline}

\text{Conservation of Energy}

\text{Velocity of the Ring}

\text{Velocity of the Solid Cylinder}

\text{Ratio of Velocities}

\text{Finding } x

\text{Conclusion}

The Sigma Insight: Rolling Motion

The Physics of Pure Rolling

Analyzing the Ring

Analyzing the Solid Cylinder

The Final Ratio

Similar Questions

Consider a situation in which a ring, a solid cylinder and a solid sphere roll down on the same inclined plane without slipping. Assume that they start rolling from rest and having identical diameter. The correct statement for this situation.

A circular disc reaches from top to bottom of an inclined plane of length $L$ . When it slips down the plane, it takes time $t_{1}$ . When it rolls down the plane, it takes time $t_{2}$ . The value of $\frac{t _{2}}{t _{1}}$ is $\frac{3}{x}$ . The value of $x$ will be …… .

The centre of a wheel rolling on a plane surface moves with a speed $v_{0}$ . A particle on the rim of the wheel at the same level as the centre will be moving at a speed $x v_{0}$ . Then, the value of $x$ is ……… .

An annular ring with inner and outer radii $R_{1}$ and $R_{2}$ is rolling without slipping with a uniform angular speed. The ratio of the forces experienced by the two particles situated on the inner and outer parts of the ring, $F_{1} / F_{2}$ is

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 $h_{sph}$ and $h_{cyl}$ on the incline. The ratio $\frac{h _{sph}}{h _{cyl}}$ is given by