Question

A person starts from rest at the top of a large frictionless spherical surface, and slides into the water below (see the drawing). At what angle theta does the person leave the surface? At what speed does the person enter the water?

Answer 1

Q2

Person starts from rest, that is initial speed is zero $V_{i}=0$

At a point where radius makes an angle $\theta$ with vertical let the speed be .

conserving energy of person

$mgR+\frac{1}{2}mV_i^2=mgR\cos{\theta}+\frac{1}{2}mV^2$

$V^2=2gR(1-\cos{\theta})$

$V=\sqrt{2gR(1-\cos{\theta})}$

At this point forces on the person are gravitational force and Normal reaction .

net force towards the center is giving the centripetal acceleration $mg\cos{\theta}-N=\frac{mV^2}{R}$

(a)

If the person is about to lose contact at this point Normal reaction is $N=0$ . That is $mg\cos{\theta}=\frac{mV^2}{R}$

Hence $V^2=gR\cos{\theta}$

from earlier equation $V^2=2gR(1-\cos{\theta})$

Using the above equations $\cos{\theta}=\frac{2}{3}$

That is the person leaves the surface at an angle $\theta=\arccos{\frac{2}{3}}=48.2^{\circ}$

(b)

Just after the person leaves contact his speed is $V=\sqrt{\frac{2gR}{3}}$

From this point his displacement to reach the water is $h=\frac{2R}{3}$

conserving energy again,

$\frac{1}{2}mV_{f}^2=\frac{2mgR}{3}+\frac{1}{2}mV^2$

$\frac{1}{2}mV_{f}^2=\frac{2mgR}{3}+\frac{mgR}{3}=mgR$

$V_{f}=\sqrt{2gR}$