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6.System of Particles and Rotational Motion
hard
A hollow spherical ball of uniform density rolls up a curved surface with an initial velocity $3\, m / s$ (as shown in figure). Maximum height with respect to the initial position covered by it will be $...........cm$.
A
$75$
B
$74$
C
$73$
D
$72$
(JEE MAIN-2023)
Solution
At highest point $KE _{ f }=0$
Initial $KE =$ Translational $KE +$ Rotational $KE$
$=\frac{1}{2} mv ^2+\frac{1}{2} I \omega^2$
In case of rolling $v = R \omega$
$=\frac{1}{2} m v^2+\frac{1}{2} \times \frac{2}{3} m R^2 \times \frac{v^2}{R^2}$
$=\frac{5}{6} m v^2$
Apply energy conservation
$KE _{ i }+ PE _{ i }= KE _{ f }+ PE _{ f }$
$\frac{5}{6} mv ^2= mgh$
$h =\frac{5}{6 \times 10} \times 9\,m =\frac{15}{20} m =75\,cm$
Standard 11
Physics
