Question

A boy is initially seated on top of hemispherical ice mound of radius R. He begins to slide down the ice with negligible initial velocity. Approximate ice as being frictionless. At what height H(in terms of R) does the boy lose contact with ice ?

Answer 1

Answer on Question#49489 - Physics - Mechanics | Kinematics | Dynamics

$$h = \frac {2 R}{3}$$

Solution

Due to the conservation law:

$$m g h + \frac {m V ^ {2}}{2} = c o n s t$$

At the begin point $V = 0$ , hence const = mgR

At the end point: $mgh_{1} + \frac{mV_{1}^{2}}{2} = mgR$ ; $V_{1} = \sqrt{2g(R - h_{1})}$

Formula of centripetal acceleration: $a = \frac{V^2}{R}$

Projections of the force acting on a boy:

Normal (to surface): $F_{n} = mgsin(\phi)$

Tangential (to surface): $F_{t} = mg\cos (\phi)$

We count angle from $x - axis$ .

Newton's law: $ma = F$ .

Boy will lose contact with surface when next condition becomes true:

$$(m a = F _ {n}) < m \frac {V ^ {2}}{R}$$

So we solve equation $mgsin(\phi) = m\frac{V^2}{R}$ for $\sin (\phi)$ .

$$m g s i n (\phi) = m \frac {V ^ {2}}{R}$$ $$g s i n (\phi) = \frac {2 g (R - h)}{R}$$ $$g s i n (\phi) = 2 g - \frac {2 g h}{R}$$ $$h = R \sin (\phi)$$ $$g s i n (\phi) = 2 g - \frac {2 g R \sin (\phi)}{R}$$ $$g s i n (\phi) = 2 g - 2 g \sin (\phi)$$ $$3 g \sin (\phi) = 2 g$$ $$\sin (\phi_ {1}) = \frac {2}{3}$$

Thus, boy will last contact with the surface at height $h = R \sin(\phi_1)$ :

$$h = \frac {2 R}{3}$$

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