Physics · Rotational Dynamics
JEE Advanced 2023 — Paper 2 — Question 21
A thin circular coin of mass 5 gm and radius is initially in a horizontal xy -plane. The coin is tossed vertically up ( +z direction) by applying an impulse of -s at a distance from its center. The coin spins about its diameter and moves along the +z direction. By the time the coin reaches back to its initial position, it completes rotations. The value of is -. [Given: The acceleration due to gravity ]
Answer: 30
Numerical answer — enter this value.
Step-by-step solution
\mathrm{J}=\mathrm{MV}_{\mathrm{cm}} …(1)
\end{gathered}$$
J. $\frac{\mathrm{R}}{2}=\frac{\mathrm{MR}^{2}}{4} \omega …(2)$
From (1) $\mathrm{V}_{\mathrm{cm}}=\mathrm{J} / \mathrm{M}$
Time when coin reaches back to its
initial position is $\mathrm{T}=\frac{2 \mathrm{~V}_{\mathrm{cm}}}{\mathrm{g}}=\frac{2 \mathrm{~J}}{\mathrm{gM}}$.
Angle rotated in time T is
$\theta=\omega \mathrm{t}=\frac{2 \mathrm{~J}}{\mathrm{MR}} \cdot \frac{2 \mathrm{~J}}{\mathrm{Mg}}$ [From (2)]
$\Rightarrow \theta=60 \pi$
$\therefore n=\frac{\theta}{2 \pi}=\frac{60 \pi}{2 \pi}=30$
Answer key and solution verified before publishing.
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- Exam
- JEE Advanced 2023
- Paper
- Paper 2
- Subject
- Physics
- Chapter
- Rotational Dynamics
- Topic
- Angular Impulse and Collisions with Rigid Bodies