Question

A solid sphere of mass $10\text{ kg}$ and radius $0.2\text{ m}$ is rolling without slipping on a horizontal floor with a velocity of $5\text{ m/s}$. Its total kinetic energy is:

• A. 125 J
• B. 175 J
• C. 250 J
• D. 150 J Correct Answer: (B) 175 J

Hint:

To solve rolling energy problems instantly, memorize the fixed total kinetic energy fractions for common symmetric objects: - Ring / Hollow Cylinder: $K = 1\,mv^2$ - Hollow Sphere: $K = \frac{5}{6}mv^2$ - Disc / Solid Cylinder: $K = \frac{3}{4}mv^2$ - Solid Sphere: $K = \frac{7}{10}mv^2$ Knowing these fractions allows you to skip writing out the moment of inertia steps entirely during a time-pressured test!

Answer 1

The Correct Option is B

Step 1: Split the energy into two parts.
A rolling sphere both slides forward and spins, so its energy is the sliding part plus the spinning part: \[ K = \tfrac{1}{2}mv^2 + \tfrac{1}{2}I\omega^2 \]

Step 2: Find the sliding (forward) energy.
With $m=10\text{ kg}$ and $v=5\text{ m/s}$: \[ K_t = \tfrac{1}{2}\times 10 \times 5^2 = 5\times 25 = 125\text{ J} \]

Step 3: Find the spinning energy.
For a solid sphere $I=\tfrac{2}{5}mR^2$ and rolling means $\omega = v/R$. The radius cancels, leaving \[ K_r = \tfrac{1}{2}\times\tfrac{2}{5}m v^2 = \tfrac{1}{5}\times 10 \times 25 = 50\text{ J} \]

Step 4: Add the two parts.
\[ K = 125 + 50 = 175\text{ J} \]
So the total kinetic energy is 175 J, which is option (B).
\[ \boxed{175\text{ J}} \]