If an object's angular velocity is increased by 10 per cent, then K.E must be increased by what percentage ?

(a) 40 %
(b) 20 %
(c) 10 %
(d) 21 %

Question

If an object's angular velocity is increased by 10 per cent, then K.E must be increased by what percentage ?

(a) 40 %
(b) 20 %
(c) 10 %
(d) 21 %

Accepted Answer

>

> [!TLDR]
> Rotational kinetic energy varies as the square of angular velocity, so a 10% rise in ω produces a $(1.1)^2 = 1.21$ times increase in KE — i.e. a 21% rise, matching option (d).
>
> The rotational kinetic energy of a body is given by:
>
> $$KE = \dfrac{1}{2}I\omega^2$$
>
> For a fixed moment of inertia $I$, $KE \propto \omega^2$.
>
> Let the original angular velocity be $\omega$, so $KE_1 = \dfrac{1}{2}I\omega^2$.
>
> If $\omega$ is increased by 10%, the new angular velocity is $\omega_2 = 1.10\,\omega$.
>
> $$KE_2 = \dfrac{1}{2}I(1.10\,\omega)^2 = \dfrac{1}{2}I\omega^2 	imes (1.10)^2 = KE_1 	imes 1.21$$
> …

Linear Quantity	Rotational Analogue
Displacement $x$	Angular displacement $\theta$
Velocity $v$	Angular velocity $\omega$
Acceleration $a$	Angular acceleration $\alpha$
Mass $m$	Moment of inertia $I$
Force $F$	Torque $\tau$
Newton's 2nd law: $F = ma$	$\tau = I\alpha$
Kinetic energy: $\frac{1}{2}mv^2$	$\frac{1}{2}I\omega^2$
Momentum: $p = mv$	Angular momentum: $L = I\omega$

Q.If an object's angular velocity is increased by 10 per cent, then K.E must be increased by what percentage ?

Concept understanding — Rotational Dynamics

Rotational Dynamics: The Physics of Spinning Things

The Core Idea

Breaking It Down

Torque — The Rotational "Push"

Moment of Inertia — The Rotational "Mass"

Angular Acceleration — How Fast Rotation Changes

The Complete Picture: Rotational Analogues

The Key Insight: Conservation of Angular Momentum

Q.If an object's angular velocity is increased by 10 per cent, then K.E must be increased by what percentage ?

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