Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
A uniform disc of radius $a$ and mass $m$ is rotating freely with an angular speed of $\omega$ in a horizontal plane about a smooth fixed vertical axis passing through its centre. A particle, also of mass $m$ is suddenly attached to the rim of the disc and starts to rotate with it. The new angular speed of this system is
PhysicsRotational MotionAP EAMCETAP EAMCET 2021 (25 Aug Shift 1)
Options:
  • A $\frac{\omega}{3}$
  • B $\frac{\omega}{6}$
  • C $\frac{\omega}{2}$
  • D $\frac{\omega}{5}$
Solution:
2228 Upvotes Verified Answer
The correct answer is: $\frac{\omega}{3}$
Given, mass of disc $=m$
Radius of disc $=a$
Angular speed $=\omega$
Moment of inertia of the disc, $I=\frac{1}{2} m a^2$
Initial angular momentum, $L_i=I \times \omega=\frac{1}{2} m a^2 \omega$
Let $\omega^{\prime}$ be the angular momentum of the disc when the particle of mass $m$ is attached at the rim.
Final angular momentum, $L_f=I \omega^{\prime}+m a v^{\prime}$
$$
\begin{aligned}
& =\frac{1}{2} m a^2 \omega^{\prime}+m a^2 \omega^{\prime} \\
& {\left[\therefore v^{\prime}=\omega^{\prime} a\right]}
\end{aligned}
$$
Applying conservation of angular momentum,
$$
\begin{aligned}
L_i & =L_f \\
\frac{1}{2} m a^2 \omega & =\frac{1}{2} m a^2 \omega^{\prime}+m a^2 \omega^{\prime} \\
\Rightarrow \quad \omega^{\prime} & =\frac{\omega}{3}
\end{aligned}
$$

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.