Physics Q89 – AP-EAMCET 2026

Home Physics Rotational MotionQuestion #426

A torque '

T

' produces angular acceleration

α

in a circular disc. If radius of disc is doubled keeping the mass constant, angular acceleration becomes

Clue 1 of 3: Rotational Dynamics

Understand the relationship between torque, moment of inertia, and angular acceleration.

🧠 Full Solution Path

Step 1: Recall the fundamental relationship between torque, moment of inertia, and angular acceleration✦ Active

The rotational equivalent of Newton's second law states that torque ( $τ$ ) is directly proportional to angular acceleration ( $α$ ) and the constant of proportionality is the moment of inertia ( $I$ ). Mathematically, this is expressed as:

τ = I α

💡 Teacher's Secret Hint

Remember that this equation is analogous to $F = m a$ in linear motion, where torque is like force, moment of inertia is like mass, and angular acceleration is like linear acceleration.

Step 2: Express the initial conditions for the circular disc○ Expand

For a solid circular disc of mass $M$ and radius $R$ , the moment of inertia is given by:

I = \frac{1}{2} M R^{2}

Given that the initial torque is $T$ and the initial angular acceleration is $α$ , we can write the initial relationship as:

T = (\frac{1}{2} M R^{2}) α (E q u a t i o n 1)

💡 Teacher's Secret Hint

Ensure you use the correct formula for the moment of inertia of a solid disc. Different shapes (e.g., ring, sphere) have different formulas.

Step 3: Determine the new moment of inertia after changing the radius○ Expand

The problem states that the mass $M$ remains constant, but the radius of the disc is doubled. Let the new radius be $R^{'}$ . Then $R^{'} = 2 R$ . The new moment of inertia, $I^{'}$ , will be:

I^{'} = \frac{1}{2} M (R^{'})^{2} = \frac{1}{2} M (2 R)^{2} = \frac{1}{2} M (4 R^{2}) = 4 (\frac{1}{2} M R^{2})

From Equation 1, we know that $I = \frac{1}{2} M R^{2}$ . So, the new moment of inertia is:

I^{'} = 4 I

💡 Teacher's Secret Hint

Squaring the radius $2 R$ gives $4 R^{2}$ , leading to a factor of 4 increase in the moment of inertia. This is a common point of error if one forgets to square the radius.

Step 4: Calculate the new angular acceleration with the constant torque○ Expand

The torque $T$ is kept constant. Let the new angular acceleration be $α^{'}$ . Using the relation $τ = I α$ for the new conditions:

T = I^{'} α^{'}

Substitute $I^{'} = 4 I$ into this equation:

T = (4 I) α^{'} (E q u a t i o n 2)

Now, equate Equation 1 and Equation 2 since the torque $T$ is the same:

I α = 4 I α^{'}

Divide both sides by $I$ (assuming $I \neq 0$ ):

α = 4 α^{'}

Solve for $α^{'}$ :

α^{'} = \frac{α}{4}

Thus, the angular acceleration becomes $α / 4$ .

💡 Teacher's Secret Hint

Remember that if torque is constant, and moment of inertia increases, then angular acceleration must decrease proportionally to keep the product constant. Since $I$ increased by a factor of 4, $α$ must decrease by a factor of 4.

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Physics Question 89 – AP-EAMCET 2026

Clue 1 of 3: Rotational Dynamics

🧠 Full Solution Path