MCQ
A small object of uniform density rolls up a curved surface with an initial velocity v'. If reaches up to a maximum height of $\frac{3 v^2}{4 g}$ with respect to the initial position. The object is
- Asolid sphere
- Bring
- Cdisc
- Dhollow sphere
✓
Answer
(c) disc
Explanation: Loss in (translational K.E. + rotational K.E.) = Gain in P.E.
$
\begin{aligned}
& \frac{1}{2} m v^2+\frac{1}{2} T \omega^2=m g h_{\max } \\
& \frac{1}{2} m v^2+\frac{1}{2} I\left(\frac{v}{R}\right)^2=m g \times \frac{3 v^2}{4 g}=\frac{3}{4} m v^2 \\
& \Rightarrow I=\frac{1}{2} m R^2
\end{aligned}
$
Explanation: Loss in (translational K.E. + rotational K.E.) = Gain in P.E.
$
\begin{aligned}
& \frac{1}{2} m v^2+\frac{1}{2} T \omega^2=m g h_{\max } \\
& \frac{1}{2} m v^2+\frac{1}{2} I\left(\frac{v}{R}\right)^2=m g \times \frac{3 v^2}{4 g}=\frac{3}{4} m v^2 \\
& \Rightarrow I=\frac{1}{2} m R^2
\end{aligned}
$
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