First slide

Dynamics of rotational motion about fixed axis

Question

A uniform solid sphere rolls on a horizontal surface at 20 m/s. It then rolls up an incline having an angle of inclination of 30^o with the Horizontal. If the friction losses are negligible, the value of height h above the ground where the ball stops is

Easy

A

14.3 m
B

28.6 m
C

57.2 m
D

9.8 m

Solution

Applying the law of conservation of energy, we have
$\frac{1}{2} {mv}^{2} + \frac{1}{2} {Iω}^{2} = mgh$
or $\frac{1}{2} {mv}^{2} + \frac{1}{2} (\frac{2}{5} {mr}^{2}) ω^{2} = mgh$
or $\frac{1}{2} {mv}^{2} + \frac{1}{5} {mv}^{2} = mgh$
$h = \frac{7}{10} \times \frac{v^{2}}{g} = \frac{7 \times (20)^{2}}{10 \times 9 \cdot 8} = 28 \cdot 6 m$

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