Physics Q90 – AP-EAMCET 2026

Home Physics Rotational MotionQuestion #425

A ring of mass

1.5 kg

and radius

35 cm

is rolling without slipping on a horizontal floor. If the translational kinetic energy of the ring is

100 J

, then its rotational kinetic energy is

Clue 1 of 3: Rolling Motion

Understand that for a body rolling without slipping, its total kinetic energy is the sum of its translational and rotational kinetic energies.

🧠 Full Solution Path

Step 1: Identify properties of a ring and rolling motion✦ Active

A ring is a hollow cylinder. Its moment of inertia about an axis passing through its center of mass and perpendicular to its plane is given by $I = m R^{2}$ , where $m$ is the mass and $R$ is the radius of the ring.

For an object rolling without slipping, the linear velocity of its center of mass $v$ is related to its angular velocity $ω$ by the equation $v = R ω$ . This implies $ω = \frac{v}{R}$ .

Step 2: Express rotational kinetic energy in terms of linear velocity○ Expand

The rotational kinetic energy ( $K E_{r o t}$ ) is given by the formula:

K E_{r o t} = \frac{1}{2} I ω^{2}

Substitute the moment of inertia for a ring ( $I = m R^{2}$ ) and the rolling condition ( $ω = \frac{v}{R}$ ) into the rotational kinetic energy formula:

K E_{r o t} = \frac{1}{2} (m R^{2}) {(\frac{v}{R})}^{2}

K E_{r o t} = \frac{1}{2} (m R^{2}) (\frac{v^{2}}{R^{2}})

K E_{r o t} = \frac{1}{2} m v^{2}

💡 Teacher's Secret Hint

Ensure that you correctly substitute the expression for $ω$ in terms of $v$ and $R$ . Notice how the $R^{2}$ terms cancel out in this specific case for a ring.

Step 3: Compare rotational kinetic energy with translational kinetic energy○ Expand

The translational kinetic energy ( $K E_{t r a n s}$ ) of the ring is given by:

K E_{t r a n s} = \frac{1}{2} m v^{2}

From Step 2, we found that for a ring rolling without slipping, $K E_{r o t} = \frac{1}{2} m v^{2}$ . Therefore, we can conclude that for a ring rolling without slipping, its rotational kinetic energy is equal to its translational kinetic energy.

K E_{r o t} = K E_{t r a n s}

💡 Teacher's Secret Hint

This relationship ( $K E_{r o t} = K E_{t r a n s}$ ) is specific to a ring. For other shapes like a solid cylinder or sphere, this ratio would be different.

Step 4: Calculate the rotational kinetic energy○ Expand

The problem states that the translational kinetic energy of the ring is $100 J$ .

Since we established that $K E_{r o t} = K E_{t r a n s}$ for a ring rolling without slipping, the rotational kinetic energy is also $100 J$ .

K E_{r o t} = 100 J

💡 Teacher's Secret Hint

Note that the mass ( $1.5 kg$ ) and radius ( $35 cm$ ) provided in the question are not necessary for solving this problem, as the relationship between $K E_{r o t}$ and $K E_{t r a n s}$ for a ring is a direct one.

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Physics Question 90 – AP-EAMCET 2026

Clue 1 of 3: Rolling Motion

🧠 Full Solution Path