A small object of uniform density rolls up a curved surface with an initial velocity ‘v’. It reaches up to a maximum heights of \(\frac{3v^2}{4g}\) with respect to the initial position. The object is :
Answer & explanation
Correct answer: option 1
From conservation of mechanical energy
\(\frac{I \omega^2}{2}+ \frac{1}{2}mv^2 = mgh \)
We know that v = ωR
\(\frac{1}{2}\)\(\frac{v^2}{R^2}\) + \(\frac{1}{2}\)mv2 = mgh
∴ h = \(\frac{3v^2}{4g}\)
\(\frac{1}{2}\)\(\frac{v^2}{R^2}\) + \(\frac{1}{2}\)mv2 = mg\(\frac{3v^2}{4g}\)
On calculating we get
I = \(\frac{mR^2}{2}\) which is for disc