Physics Q90 – AP-EAMCET 2026

Home Physics Rotational MotionQuestion #427

Hollow sphere rolls on an inclined plane of height

h

without slipping. If it starts from rest, its speed at the bottom is

Clue 1 of 3: Conservation of Energy

The total mechanical energy of the hollow sphere is conserved as it rolls down the inclined plane without slipping.

🧠 Full Solution Path

Step 1: Identify Initial and Final Energy States✦ Active

At the top of the inclined plane (initial state), the hollow sphere starts from rest at a height $h$ . Therefore, its initial translational kinetic energy ( $K E_{t, i n i t i a l}$ ) and rotational kinetic energy ( $K E_{r, i n i t i a l}$ ) are zero. It possesses only gravitational potential energy ( $P E_{i n i t i a l}$ ). At the bottom of the inclined plane (final state), the sphere has reached a height of zero (our reference point), so its final potential energy ( $P E_{f i n a l}$ ) is zero. It possesses both translational kinetic energy ( $K E_{t, f i n a l}$ ) and rotational kinetic energy ( $K E_{r, f i n a l}$ ). Since the sphere rolls without slipping, mechanical energy is conserved.

P E_{i n i t i a l} = M g h

K E_{t, i n i t i a l} = 0

K E_{r, i n i t i a l} = 0

P E_{f i n a l} = 0

💡 Teacher's Secret Hint

Remember that for an object rolling without slipping, there's no energy loss due to friction, so mechanical energy is conserved.

Step 2: Apply Conservation of Mechanical Energy○ Expand

According to the principle of conservation of mechanical energy, the total energy at the initial position (top) must equal the total energy at the final position (bottom).

P E_{i n i t i a l} + K E_{t, i n i t i a l} + K E_{r, i n i t i a l} = P E_{f i n a l} + K E_{t, f i n a l} + K E_{r, f i n a l}

Substituting the energy values from Step 1, we get:

M g h + 0 + 0 = 0 + \frac{1}{2} M v^{2} + \frac{1}{2} I ω^{2}

M g h = \frac{1}{2} M v^{2} + \frac{1}{2} I ω^{2}

💡 Teacher's Secret Hint

Always break down total kinetic energy into translational and rotational components for rolling objects.

Step 3: Substitute Moment of Inertia and Rolling Condition○ Expand

For a hollow sphere of mass $M$ and radius $R$ , its moment of inertia ( $I$ ) about an axis passing through its center is given by:

I = \frac{2}{3} M R^{2}

For an object rolling without slipping, the linear velocity ( $v$ ) of its center of mass is related to its angular velocity ( $ω$ ) by:

v = R ω ⟹ ω = \frac{v}{R}

Substitute these expressions for $I$ and $ω$ into the energy conservation equation from Step 2:

M g h = \frac{1}{2} M v^{2} + \frac{1}{2} (\frac{2}{3} M R^{2}) {(\frac{v}{R})}^{2}

M g h = \frac{1}{2} M v^{2} + \frac{1}{2} (\frac{2}{3} M R^{2}) (\frac{v^{2}}{R^{2}})

The $R^{2}$ terms cancel out:

M g h = \frac{1}{2} M v^{2} + \frac{1}{3} M v^{2}

💡 Teacher's Secret Hint

Ensure you use the correct moment of inertia for the object (hollow sphere, solid sphere, cylinder, etc.) and correctly apply the rolling without slipping condition, $v = R ω$ .

Step 4: Solve for the Final Velocity○ Expand

Now, we can simplify the equation and solve for $v$ . First, cancel $M$ from all terms:

g h = \frac{1}{2} v^{2} + \frac{1}{3} v^{2}

Combine the terms with $v^{2}$ :

g h = v^{2} (\frac{1}{2} + \frac{1}{3})

g h = v^{2} (\frac{3 + 2}{6})

g h = v^{2} (\frac{5}{6})

Solve for $v^{2}$ :

v^{2} = \frac{6 g h}{5}

Finally, take the square root to find $v$ :

v = \sqrt{\frac{6 g h}{5}}

💡 Teacher's Secret Hint

Double-check your fraction addition to avoid common arithmetic errors. The final answer should have units of velocity (m/s) if $g$ is in m/s $^{2}$ and $h$ in meters.

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Physics Question 90 – AP-EAMCET 2026

Clue 1 of 3: Conservation of Energy

🧠 Full Solution Path