AIIMS 2007 Physics Rolling Motion MCQ Question

To determine the total kinetic energy of a solid sphere rolling without slipping, we need to consider both its translational and rotational kinetic energies.

Step 1: Understanding the components of kinetic energy

1. Translational Kinetic Energy (TKE): This is the energy due to the motion of the center of mass of the sphere and is given by the formula:

   $$TKE = \frac{1}{2} mv^2$$

   where $m$ is the mass and $v$ is the linear velocity.

2. Rotational Kinetic Energy (RKE): This is the energy due to the rotation of the sphere around its center of mass and is given by the formula:

   $$RKE = \frac{1}{2} I \omega^2$$

   where $I$ is the moment of inertia and $\omega$ is the angular velocity.

Step 2: Calculate Translational Kinetic Energy

Given:

• Mass, $m = 1 \, \text{kg}$
• Radius, $r = 0.1 \, \text{m}$
• Velocity, $v = 1 \, \text{m/s}$

Calculating the translational kinetic energy:

$$TKE = \frac{1}{2} mv^2 = \frac{1}{2} \times 1 \times (1)^2 = \frac{1}{2} \, \text{J}$$

Step 3: Calculate the moment of inertia for a solid sphere

The moment of inertia $I$ for a solid sphere is given by the formula:

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

Substituting the values:

$$I = \frac{2}{5} \times 1 \times (0.1)^2 = \frac{2}{5} \times 1 \times 0.01 = \frac{2}{500} = \frac{1}{250} \, \text{kg m}^2$$

Step 4: Relate angular velocity to linear velocity

Since the sphere rolls without slipping, we have the relation:

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

Substituting the values:

$$\omega = \frac{1}{0.1} = 10 \, \text{rad/s}$$

Step 5: Calculate Rotational Kinetic Energy

Now we can calculate the rotational kinetic energy:

$$RKE = \frac{1}{2} I \omega^2 = \frac{1}{2} \times \frac{1}{250} \times (10)^2$$ $$= \frac{1}{2} \times \frac{1}{250} \times 100 = \frac{50}{250} = \frac{1}{5} \, \text{J}$$

Step 6: Total Kinetic Energy

Now, we sum the translational and rotational kinetic energies to find the total kinetic energy:

$$KE_{total} = TKE + RKE = \frac{1}{2} + \frac{1}{5}$$

To add these fractions, we need a common denominator. The least common multiple of 2 and 5 is 10:

$$KE_{total} = \frac{5}{10} + \frac{2}{10} = \frac{7}{10} \, \text{J}$$

Conclusion

Thus, the total kinetic energy of the rolling solid sphere is:

$$\text{Total Kinetic Energy} = \frac{7}{10} \, \text{J}$$

Correct Answer: C) $\frac{7}{10} \, \text{J}$

Clarification of Other Options

• A) $\frac{7}{5} \, \text{J}$: This value is too high, exceeding the calculated total kinetic energy.
• B) $\frac{2}{5} \, \text{J}$: This only considers part of the translational or rotational energy, not both.
• D) $1 \, \text{J}$: This also exceeds the total kinetic energy calculated.

In summary, the correct total kinetic energy of the solid sphere is $\frac{7}{10} \, \text{J}$, which accounts for both its translational and rotational motion.