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: Understanding the Concept:
When a rigid body undergoes pure rolling motion (rolling without slipping) across a flat surface, its kinetic movement is a simultaneous combination of two distinct mechanical behaviors: 1. Translational Kinetic Energy ($K_t$): The energy due to the linear movement of the body's center of mass moving forward at a velocity $v$. 2. Rotational Kinetic Energy ($K_r$): The energy due to the body spinning around its central axis at an angular velocity $\omega$. The total kinetic energy ($K_{\text{total}}$) of the rolling body is simply the direct sum of these two separate energy components. Under pure rolling conditions, the linear velocity of the center of mass and the angular spin velocity are directly locked by the kinematic boundary condition: $v = R\omega$.

Step 2: Key Formula or Approach:
1. Total Kinetic Energy equation: $$ K_{\text{total}} = K_t + K_r = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 $$ 2. Moment of Inertia ($I$): For a uniform solid sphere rotating around its central diameter axis, the moment of inertia is: $$ I = \frac{2}{5}mR^2 $$ 3. Pure Rolling Substitute: Replacing $\omega$ with $\frac{v}{R}$ and substituting $I$ into the primary energy equation gives: $$ K_{\text{total}} = \frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{2}{5}mR^2\right)\left(\frac{v}{R}\right)^2 $$ $$ K_{\text{total}} = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \left(\frac{1}{2} + \frac{1}{5}\right)mv^2 = \frac{7}{10}mv^2 $$

Step 3: Detailed Explanation:
Let's collect the given values from the question text: - Mass of the solid sphere ($m$) = $10\text{ kg}$ - Radius of the sphere ($R$) = $0.2\text{ m}$ - Linear center-of-mass velocity ($v$) = $5\text{ m/s}$ Notice that because our simplified expression $\frac{7}{10}mv^2$ contains only mass and linear velocity, the specific radius parameter ($R = 0.2\text{ m}$) is extra information that is not required for the calculation! Let's plug our values directly into the rolling energy equation: $$ K_{\text{total}} = \frac{7}{10} \times 10\text{ kg} \times (5\text{ m/s})^2 $$ Simplify the expression step-by-step: 1. Cancel out the factor of 10 in the numerator and denominator: $$ K_{\text{total}} = 7 \times (5)^2 $$ 2. Square the velocity term: $$ 5^2 = 25 $$ 3. Multiply the remaining terms together: $$ K_{\text{total}} = 7 \times 25 = 175\text{ J} $$ The total kinetic energy is exactly $175\text{ J}$, matching option (B).

Step 4: Final Answer:
The total kinetic energy of the rolling solid sphere is $175\text{ J}$.