A uniform sphere of mass 500 g rolls without slipping on a plane horizontal surface with its centre

Q: Why do we have to add two different kinetic energies?

Because the sphere is doing two distinct motions simultaneously. Its center of mass is moving forward linearly (translational energy), and the mass of the sphere is rotating around that center (rotational energy). Both require energy to maintain.

Q: What if the problem didn't say 'without slipping'?

If the sphere was slipping, the relationship $v = \omega r$ would no longer hold true. You would need more information about the friction and the forces acting on the sphere to determine its rotational speed independently of its linear speed.

Q: Why did the radius of the sphere disappear from the final formula?

This is a beautiful mathematical cancellation! The moment of inertia has an $r^2$ in the numerator, and the $\omega^2$ term introduces an $r^2$ in the denominator. They cancel out, meaning the total kinetic energy of a rolling sphere depends only on its mass and linear velocity, not its size.

Q: I got the wrong power of 10 in my answer. What happened?

You likely forgot to convert the units. The mass was given in grams (must be converted to kg by dividing by 1000) and the velocity in cm/s (must be converted to m/s by dividing by 100). Always convert to SI units first!

Q: Would the answer be different if it was a hollow sphere?

Yes! A hollow sphere has a different mass distribution, meaning its moment of inertia is $\frac{2}{3}mr^2$ instead of $\frac{2}{5}mr^2$. This would change the final fraction from $\frac{7}{10}$ to $\frac{5}{6}$.

Question

A uniform sphere of mass $500	ext{ g}$ rolls without slipping on a plane horizontal surface with its centre moving at a speed of $5.00	ext{ cm/s}$. Its kinetic energy is

Options:

(A) $6.25 	imes 10^{-4}	ext{ J}$

(B) $1.13 	imes 10^{-3}	ext{ J}$

(C) $8.75 	imes 10^{-4}	ext{ J}$

(D) $8.75 	imes 10^{-3}	ext{ J}$

Sigma Percentile · Accepted Answer

### The Dual Nature of Rolling Energy

Imagine a bowling ball rolling smoothly down a lane. It isn't just sliding forward; it's also spinning. In physics, we call this **pure rolling** (or rolling without slipping). Because the object is doing two things at once—translating through space and rotating about its center of mass—it possesses two distinct forms of kinetic energy.

The total kinetic energy ($K_{total}$) is the sum of the translational kinetic energy ($K_{trans}$) and the rotational kinetic energy ($K_{rot}$):

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

### Setting Up the Equation

To solve our problem, we need to express the rotational kinetic energy in terms of the linear velocity $v$. We are dealing with a **uniform solid sphere**. The moment of inertia ($I$) for a solid sphere about its central axis is:

$$I = \frac{2}{5}mr^2$$

Furthermore, the condition for pure rolling links the angular velocity ($\omega$) to the linear velocity ($v$) of the center of mass:

$$\omega = \frac{v}{r}$$

Let's substitute these into our total energy equation:

$$K_{total} = \frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{2}{5}mr^2ight)\left(\frac{v}{r}ight)^2$$

### The Elegance of Cancellation

Notice what happens to the radius $r$. When we square the angular velocity term, we get $\frac{v^2}{r^2}$. The $r^2$ in the numerator from the moment of inertia perfectly cancels the $r^2$ in the denominator!

$$K_{total} = \frac{1}{2}mv^2 + \frac{1}{5}mv^2$$

Adding these fractions together gives us a beautiful, simplified formula for the total kinetic energy of any rolling solid sphere, regardless of its radius:

$$K_{total} = \frac{7}{10}mv^2$$

### Final Calculation

Before we plug in the numbers, we must ensure all units are in the standard SI system to get our answer in Joules.

Mass: $m = 500 	ext{ g} = 0.5 	ext{ kg}$
Velocity: $v = 5.00 	ext{ cm/s} = 0.05 	ext{ m/s}$

Now, substitute these values into our simplified formula:

$$K_{total} = \frac{7}{10} 	imes 0.5 	imes (0.05)^2$$

$$K_{total} = 0.7 	imes 0.5 	imes 0.0025$$

$$K_{total} = 0.35 	imes 0.0025 = 0.000875 	ext{ J}$$

Expressing this in scientific notation, we get our final answer:

**$$K_{total} = 8.75 	imes 10^{-4} 	ext{ J}$$**

Animated Solution for Physics - Rotational Motion: A uniform sphere of mass $500 g$ rolls without slipping on a plane horizontal surface with its centre moving at a speed of $5.00 cm/s$ . Its kinetic energy is

Select Answer:

Visualized Solution

$Visualizing the Rolling Sphere$

$Total Kinetic Energy of a Rolling Body$

$Substituting Moment of Inertia and Angular Velocity$

$Simplifying the Expression$

$Final Calculation$

$What if it was a hollow sphere?$

The Sigma Insight: Rolling Motion

The Dual Nature of Rolling Energy

Setting Up the Equation

The Elegance of Cancellation

Final Calculation

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Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Rolling Sphere

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Total Kinetic Energy of a Rolling Body

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Moment of Inertia and Angular Velocity

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } What if it was a hollow sphere?