A body of mass m = 2 kg is at the top of a smooth…
A body of mass $m = 2\ \mathrm{kg}$ is at the top of a smooth hemispherical dome of height (radius) $R = H = 10\ \mathrm{m}$. (a) What is the smallest horizontal speed $v_0$ with which it leaves the dome immediately, without sliding on its surface? (b) If it is given half that speed, where does it leave the dome?
Explanation of Method
A body sliding off a hemispherical dome.
Step 1: (a) Leaving at once
At the top, gravity alone must be enough to supply the centripetal force (normal force zero): \[\frac{mv_0^2}{R} = mg \;\Longrightarrow\; v_0 = \sqrt{gR} = \sqrt{98.1}.\]
\[v_0 \approx 9.9\ \mathrm{m/s}\]
Step 2: (b) Starting with $v_0/2$
At angle $\theta$ from the top, the body has dropped $R(1 - \cos\theta)$: $v^2 = \frac{gR}{4} + 2gR\,(1 - \cos\theta)$.
It leaves where the normal force vanishes, $v^2 = gR\cos\theta$: $gR\cos\theta = \frac{gR}{4} + 2gR - 2gR\cos\theta$.
\[ 3\cos\theta = \frac94. \]
\[\begin{gathered} \cos\theta = \tfrac34, \quad \theta \approx 41.4^\circ \\ 2.5\ \mathrm{m} \text{ below the top} \end{gathered}\]
Check
With $v_0 = 0$ this gives the familiar $\cos\theta = 2/3$ (48.2°); a push makes it leave earlier $\checkmark$
Result
\[\boxed{\,v_0 \approx 9.9\ \mathrm{m/s}\,}\]
\[\boxed{\,\begin{gathered} \cos\theta = \tfrac34, \quad \theta \approx 41.4^\circ \\ 2.5\ \mathrm{m} \text{ below the top} \end{gathered}\,}\]
Remark
About the original solution. Part (a) is correct. In (b) the energy equation was set up wrongly: it put the potential energy $mgH$ at the top and $mgH(1 - \cos\theta)$ at the leaving point, i.e. it measured heights from different levels. That gave $\cos\theta = -1/4$ (104°), a point below the dome's base, which is impossible. The correct result is $\cos\theta = 3/4$, $\theta \approx 41.4^\circ$, 2.5 m below the top.