A stationary horizontal disc is free to rotate about its axis. When a torque is applied on it, its kinetic energy as a function of $\theta$, where $\theta$ is the angle by which it has rotated, is given as $k\theta^2$. If its moment of inertia is $I$, then the angular acceleration of the disc is Options: (A) $\frac{k}{2I}\theta$ (B) $\frac{k}{I}\theta$ (C) $\frac{k}{4I}\theta$ (D) $\frac{2k}{I}\theta$

This problem is a beautiful exercise in connecting energy concepts with rotational kinematics. Let's break down the journey from kinetic energy to angular acceleration. ### Analyzing the Setup We are given a horizontal disc that is free to rotate. A torque is applied, causing it to accelerate. The problem provides a unique piece of information: the kinetic energy $K$ of the disc is a function of its angular displacement $\theta$, given by the relation: $$K = k\theta^2$$ We also know the standard formula for the rotational kinetic energy of a rigid body: $$K = \frac{1}{2}I\omega^2$$ where $I$ is the moment of inertia and $\omega$ is the angular velocity. ### The Master Equation Since both expressions represent the same physical quantity (the kinetic energy of the disc), we can equate them: $$\frac{1}{2}I\omega^2 = k\theta^2$$ Our goal is to find the angular acceleration, $\alpha$. To do this, we first need to isolate a kinematic variable. Let's solve for $\omega^2$: $$\omega^2 = \frac{2k}{I}\theta^2$$ ### The Kinematic Trick Now, how do we get from $\omega$ to $\alpha$? We could take the square root to find $\omega$ as a function of $\theta$, and then use the chain rule. However, there is a much more elegant mathematical trick. Recall the rotational analog of the kinematic equation $a = v \frac{dv}{dx}$. For rotational motion, this is: $$\alpha = \omega \frac{d\omega}{d\theta}$$ Notice that if we differentiate $\omega^2$ with respect to $\theta$ using the chain rule, we get exactly what we need: $$\frac{d}{d\theta}(\omega^2) = 2\omega \frac{d\omega}{d\theta} = 2\alpha$$ ### Final Calculation Let's apply this trick by differentiating our $\omega^2$ equation with respect to $\theta$ on both sides: $$\frac{d}{d\theta}(\omega^2) = \frac{d}{d\theta}\left(\frac{2k}{I}\theta^2\right)$$ Applying the derivatives: $$2\omega \frac{d\omega}{d\theta} = \frac{2k}{I}(2\theta)$$ We can cancel the factor of $2$ from both sides: $$\omega \frac{d\omega}{d\theta} = \frac{2k}{I}\theta$$ Since $\omega \frac{d\omega}{d\theta}$ is exactly our angular acceleration $\alpha$, we arrive at our final elegant result: $$\alpha = \frac{2k}{I}\theta$$ This tells us that the angular acceleration is directly proportional to the angular displacement, which means the torque applied is not constant, but increases as the disc rotates further!

Why did we use $\alpha = \omega \frac{d\omega}{d\theta}$ instead of $\alpha = \frac{d\omega}{dt}$?

Because the kinetic energy (and thus $\omega$) is given as a function of angular displacement $\theta$, not time $t$. Using the chain rule relation $\alpha = \omega \frac{d\omega}{d\theta}$ allows us to differentiate directly with respect to the variable we have, avoiding complex time-parameterizations.

Could I have taken the square root first to find $\omega = \sqrt{\frac{2k}{I}}\theta$?

Yes, absolutely! If you take the square root, you get $\omega = \sqrt{\frac{2k}{I}}\theta$. Then, you can differentiate this with respect to time: $\frac{d\omega}{dt} = \sqrt{\frac{2k}{I}} \frac{d\theta}{dt}$. Since $\frac{d\theta}{dt} = \omega$, you substitute $\omega$ back in to get $\alpha = \sqrt{\frac{2k}{I}} \left(\sqrt{\frac{2k}{I}}\theta\right) = \frac{2k}{I}\theta$. Both methods are perfectly valid, but differentiating $\omega^2$ directly is often faster and less prone to algebraic errors.

What does it mean physically if $\alpha$ is proportional to $\theta$?

It means the motion is not uniformly accelerated. As the disc rotates and $\theta$ increases, the angular acceleration $\alpha$ also increases. Since $\tau = I\alpha$, this implies that the applied torque is also increasing proportionally with the angle of rotation.

JEE Main 2019, 9 April Shift-I

LEVELJEE Main

Animated Solution for Physics - Rotational Motion: A stationary horizontal disc is free to rotate about its axis. When a torque is applied on it, its kinetic energy as a function of $θ$ , where $θ$ is the angle by which it has rotated, is given as $k θ^{2}$ . If its moment of inertia is $I$ , then the angular acceleration of the disc is

Q: Why did we cancel the 2 in the differentiation step?

When we differentiated $\omega^2$, we got $2\omega \frac{d\omega}{d\theta}$. When we differentiated $\theta^2$, we got $2\theta$. Since there is a factor of 2 on both sides of the equation, we can divide the entire equation by 2 to simplify it.

Q: Is $k$ a spring constant here?

Not necessarily. The problem just states $k$ is a constant in the given mathematical function for kinetic energy. While the math looks similar to the potential energy of a torsional spring ($U = \frac{1}{2}C\theta^2$), here it represents kinetic energy, so it's just a generic proportionality constant.

Select Answer:

Visualized Solution

\text{Given Information}

$K = k θ^{2}$

\text{Kinetic Energy Formula}

$K = \frac{1}{2} I ω^{2}$

\text{Equating Kinetic Energies}

$\frac{1}{2} I ω^{2} = k θ^{2}$

\text{Solving for } \omega^2

$ω^{2} = \frac{2 k}{I} θ^{2}$

\text{Relation between } \alpha, \omega \text{ and } \theta

$α = ω \frac{d ω}{d θ}$

\text{Differentiating } \omega^2

$\frac{d}{d θ} (ω^{2}) = \frac{d}{d θ} (\frac{2 k}{I} θ^{2})$

\text{Applying Chain Rule}

$2 ω \frac{d ω}{d θ} = \frac{2 k}{I} (2 θ)$

\text{Final Answer}

$ω \frac{d ω}{d θ} = \frac{2 k}{I} θ$
$α = \frac{2 k}{I} θ$

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The Sigma Insight: Dynamics of Rigid Body Rotation

Solution Diagram

This problem is a beautiful exercise in connecting energy concepts with rotational kinematics. Let's break down the journey from kinetic energy to angular acceleration.

Analyzing the Setup We are given a horizontal disc that is free to rotate

A torque is applied, causing it to accelerate. The problem provides a unique piece of information: the kinetic energy

K

of the disc is a function of its angular displacement

θ

, given by the relation:

K = k θ^{2}

We also know the standard formula for the rotational kinetic energy of a rigid body:

K = \frac{1}{2} I ω^{2}

where

I

is the moment of inertia and

ω

is the angular velocity.

The Master Equation

Since both expressions represent the same physical quantity (the kinetic energy of the disc), we can equate them:

\frac{1}{2} I ω^{2} = k θ^{2}

Our goal is to find the angular acceleration,

α

. To do this, we first need to isolate a kinematic variable. Let's solve for

ω^{2}

ω^{2} = \frac{2 k}{I} θ^{2}

The Kinematic Trick Now, how do we get from $ω$ to $α$ ? We could take the square root to find $ω$ as a function of $θ$ , and then use the chain rule

However, there is a much more elegant mathematical trick.

Recall the rotational analog of the kinematic equation

a = v \frac{d v}{d x}

. For rotational motion, this is:

α = ω \frac{d ω}{d θ}

Notice that if we differentiate

ω^{2}

with respect to

θ

using the chain rule, we get exactly what we need:

\frac{d}{d θ} (ω^{2}) = 2 ω \frac{d ω}{d θ} = 2 α

Final Calculation

Let's apply this trick by differentiating our

ω^{2}

equation with respect to

θ

on both sides:

\frac{d}{d θ} (ω^{2}) = \frac{d}{d θ} (\frac{2 k}{I} θ^{2})

Applying the derivatives:

2 ω \frac{d ω}{d θ} = \frac{2 k}{I} (2 θ)

We can cancel the factor of

2

from both sides:

ω \frac{d ω}{d θ} = \frac{2 k}{I} θ

Since

ω \frac{d ω}{d θ}

is exactly our angular acceleration

α

, we arrive at our final elegant result:

α = \frac{2 k}{I} θ

This tells us that the angular acceleration is directly proportional to the angular displacement, which means the torque applied is not constant, but increases as the disc rotates further!

Common Traps & FAQs

Why did we use

α = ω \frac{d ω}{d θ}

instead of

α = \frac{d ω}{d t}

Because the kinetic energy (and thus

ω

) is given as a function of angular displacement

θ

, not time

t

. Using the chain rule relation

α = ω \frac{d ω}{d θ}

allows us to differentiate directly with respect to the variable we have, avoiding complex time-parameterizations.

Could I have taken the square root first to find

ω = \frac{2 k}{I} θ

Yes, absolutely! If you take the square root, you get

ω = \frac{2 k}{I} θ

. Then, you can differentiate this with respect to time:

\frac{d ω}{d t} = \frac{2 k}{I} \frac{d θ}{d t}

. Since

\frac{d θ}{d t} = ω

, you substitute

ω

back in to get

α = \frac{2 k}{I} (\frac{2 k}{I} θ) = \frac{2 k}{I} θ

. Both methods are perfectly valid, but differentiating

ω^{2}

directly is often faster and less prone to algebraic errors.

What does it mean physically if

α

is proportional to

θ

It means the motion is not uniformly accelerated. As the disc rotates and

θ

increases, the angular acceleration

α

also increases. Since

τ = I α

, this implies that the applied torque is also increasing proportionally with the angle of rotation.

Why did we cancel the 2 in the differentiation step?

When we differentiated

ω^{2}

, we got

2 ω \frac{d ω}{d θ}

. When we differentiated

θ^{2}

, we got

2 θ

. Since there is a factor of 2 on both sides of the equation, we can divide the entire equation by 2 to simplify it.

k

a spring constant here?

Not necessarily. The problem just states

k

is a constant in the given mathematical function for kinetic energy. While the math looks similar to the potential energy of a torsional spring (

U = \frac{1}{2} C θ^{2}

), here it represents kinetic energy, so it's just a generic proportionality constant.