Question

A solid sphere of mass 2 kg and radius 10 cm is rolling down on an inclined plane without slipping. If the rotational kinetic energy of the sphere is 40 J, then its total kinetic energy is

• A. 100 J
• B. 80 J
• C. 140 J
• D. 200 J

Answer 1

The Correct Option is C

Step 1: Identify the given information and the goal.
We are given: \begin{itemize} \item Object: Solid sphere rolling down an inclined plane without slipping. \item Mass ($m$) = 2 kg (This information is not required for the solution as shown below). \item Radius ($R$) = 10 cm (This information is not required for the solution as shown below). \item Rotational kinetic energy ($KE_{rot}$) = 40 J. \end{itemize} We need to find the total kinetic energy ($KE_{total}$) of the sphere. Step 2: Recall the formulas for kinetic energies in rolling motion.
For a body rolling without slipping, the total kinetic energy is the sum of its translational kinetic energy ($KE_{trans}$) and rotational kinetic energy ($KE_{rot}$): $KE_{total} = KE_{trans} + KE_{rot}$ The translational kinetic energy is given by: $KE_{trans} = \frac{1}{2}mv^2$ where $m$ is the mass and $v$ is the translational velocity of the center of mass. The rotational kinetic energy is given by: $KE_{rot} = \frac{1}{2}I\omega^2$ where $I$ is the moment of inertia and $\omega$ is the angular velocity. For rolling without slipping, there is a relationship between translational and angular velocity: $v = R\omega \implies \omega = \frac{v}{R}$ For a solid sphere, the moment of inertia about an axis passing through its center is: $I = \frac{2}{5}mR^2$ Step 3: Establish the relationship between translational and rotational kinetic energy for a solid sphere.
Substitute the expression for $I$ and $\omega$ into the rotational kinetic energy formula: $KE_{rot} = \frac{1}{2} \left(\frac{2}{5}mR^2\right) \left(\frac{v}{R}\right)^2$ $KE_{rot} = \frac{1}{2} \times \frac{2}{5}mR^2 \times \frac{v^2}{R^2}$ $KE_{rot} = \frac{1}{5}mv^2$ Now, compare this with the translational kinetic energy: $KE_{trans} = \frac{1}{2}mv^2$. We can see that $\frac{1}{5}mv^2 = \frac{2}{5} \left(\frac{1}{2}mv^2\right) = \frac{2}{5}KE_{trans}$. So, $KE_{rot} = \frac{2}{5}KE_{trans}$. Alternatively, we can express $KE_{trans}$ in terms of $KE_{rot}$: $KE_{trans} = \frac{5}{2}KE_{rot}$ Step 4: Calculate the translational kinetic energy.
Given $KE_{rot} = 40 \text{ J}$. Using the relationship derived: $KE_{trans} = \frac{5}{2} \times 40 \text{ J}$ $KE_{trans} = 5 \times 20 \text{ J}$ $KE_{trans} = 100 \text{ J}$ Step 5: Calculate the total kinetic energy.
$KE_{total} = KE_{trans} + KE_{rot}$ $KE_{total} = 100 \text{ J} + 40 \text{ J}$ $KE_{total} = 140 \text{ J}$ The final answer matches option (3). The final answer is $\boxed{\text{140 J}}$.