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A small object of uniform density rolls up a curved surface with an initial velocity $v$. It reaches up to a maximum height of $\frac{3 v^2}{4 g}$ with respect to the initial$$
\text { position. The object is }
$$
Options:
\text { position. The object is }
$$
Solution:
2075 Upvotes
Verified Answer
The correct answer is:
disc
disc
$$
\begin{aligned}
\frac{1}{2} m v^2+\frac{1}{2} I\left(\frac{v}{R}\right)^2 & =m g\left(\frac{3 v^2}{4 g}\right) \\
I & =\frac{1}{2} m R^2
\end{aligned}
$$
So, body is disc.
The correct option is (d).
\begin{aligned}
\frac{1}{2} m v^2+\frac{1}{2} I\left(\frac{v}{R}\right)^2 & =m g\left(\frac{3 v^2}{4 g}\right) \\
I & =\frac{1}{2} m R^2
\end{aligned}
$$
So, body is disc.
The correct option is (d).
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