Problem 3: Break-off Angle for Disc on Hemisphere
1. Velocity Analysis
Let $M$ be the mass of the bowl and $m$ be the mass of the disc. Given $M=m$.
Let $V_b$ be the velocity of the bowl (horizontal) and $v_{rel}$ be the relative velocity of the disc sliding down the bowl surface at angle $\theta$.
The velocity of the disc in the lab frame, $\vec{v}_d$, is the vector sum of the bowl’s velocity and the relative velocity:
- Horizontal component: $v_{dx} = v_{rel} \cos \theta – V_b$ (assuming bowl moves left, disc horizontal component right)
- Vertical component: $v_{dy} = -v_{rel} \sin \theta$
Conservation of Momentum (Horizontal):
$$m(v_{rel} \cos \theta – V_b) – M V_b = 0$$
Since $M=m$:
$$v_{rel} \cos \theta – V_b = V_b \implies V_b = \frac{v_{rel} \cos \theta}{2}$$
Notice that the horizontal velocity of the disc is $v_{dx} = V_b$.
2. Maximization Condition
The disc breaks contact when the normal force becomes zero. This coincides with the moment the horizontal velocity of the disc reaches its maximum value. Afterward, it becomes a projectile with constant horizontal velocity.
We need to express $v_{dx}^2$ in terms of $\theta$ and maximize it.
Conservation of Energy:
$$mgR(1 – \cos \theta) = \frac{1}{2}m v_d^2 + \frac{1}{2}M V_b^2$$
Where $v_d^2 = v_{dx}^2 + v_{dy}^2 = V_b^2 + (v_{rel} \sin \theta)^2$. Substituting this:
$$mgR(1 – \cos \theta) = \frac{1}{2}m [V_b^2 + v_{rel}^2 \sin^2 \theta] + \frac{1}{2}m V_b^2$$
$$gR(1 – \cos \theta) = V_b^2 + \frac{1}{2} v_{rel}^2 \sin^2 \theta$$
Substitute $v_{rel} = \frac{2 V_b}{\cos \theta}$:
$$gR(1 – \cos \theta) = V_b^2 + \frac{1}{2} \left( \frac{4 V_b^2}{\cos^2 \theta} \right) \sin^2 \theta$$
$$gR(1 – \cos \theta) = V_b^2 \left( 1 + 2 \tan^2 \theta \right)$$
We want to maximize $v_{dx} = V_b$. Let $u = \cos \theta$.
$$V_b^2 = \frac{gR(1 – u)}{1 + 2 \frac{1-u^2}{u^2}} = \frac{gR(1 – u) u^2}{u^2 + 2 – 2u^2} = \frac{gR(1 – u) u^2}{2 – u^2}$$
3. Calculating the Critical Angle
To find the maximum, we differentiate $y = \frac{(1-u)u^2}{2-u^2}$ with respect to $u$ and set it to zero.
$$y = \frac{u^2 – u^3}{2 – u^2}$$
$$y’ = \frac{(2 – u^2)(2u – 3u^2) – (u^2 – u^3)(-2u)}{(2 – u^2)^2} = 0$$
Numerator must be zero:
$$(2 – u^2)(2u – 3u^2) + 2u(u^2 – u^3) = 0$$
$$u [ (2 – u^2)(2 – 3u) + 2(u^2 – u^3) ] = 0$$
Since $u = \cos \theta \neq 0$:
$$(4 – 6u – 2u^2 + 3u^3) + (2u^2 – 2u^3) = 0$$
$$u^3 – 6u + 4 = 0$$
We need to solve $u^3 – 6u + 4 = 0$. By inspection, or checking the proposed answer $\cos \theta = \sqrt{3} – 1$:
Let $u = \sqrt{3} – 1$. Then $u^3 = (\sqrt{3}-1)^3 = 3\sqrt{3} – 9 + 3\sqrt{3} – 1 = 6\sqrt{3} – 10$.
Substitute into equation:
$$(6\sqrt{3} – 10) – 6(\sqrt{3} – 1) + 4$$
$$6\sqrt{3} – 10 – 6\sqrt{3} + 6 + 4 = 0$$
The equation holds. Thus, the solution is:
$$\theta = \cos^{-1}(\sqrt{3} – 1)$$
1. Velocity Analysis
Let $M$ be the mass of the bowl and $m$ be the mass of the disc. Given $M=m$.
Let $V_b$ be the velocity of the bowl (horizontal) and $v_{rel}$ be the relative velocity of the disc sliding down the bowl surface at angle $\theta$.
The velocity of the disc in the lab frame, $\vec{v}_d$, is the vector sum of the bowl’s velocity and the relative velocity:
- Horizontal component: $v_{dx} = v_{rel} \cos \theta – V_b$ (assuming bowl moves left, disc horizontal component right)
- Vertical component: $v_{dy} = -v_{rel} \sin \theta$
Conservation of Momentum (Horizontal):
$$m(v_{rel} \cos \theta – V_b) – M V_b = 0$$Since $M=m$:
$$v_{rel} \cos \theta – V_b = V_b \implies V_b = \frac{v_{rel} \cos \theta}{2}$$Notice that the horizontal velocity of the disc is $v_{dx} = V_b$.
2. Maximization Condition
The disc breaks contact when the normal force becomes zero. This coincides with the moment the horizontal velocity of the disc reaches its maximum value. Afterward, it becomes a projectile with constant horizontal velocity.
We need to express $v_{dx}^2$ in terms of $\theta$ and maximize it.
Conservation of Energy:
$$mgR(1 – \cos \theta) = \frac{1}{2}m v_d^2 + \frac{1}{2}M V_b^2$$Where $v_d^2 = v_{dx}^2 + v_{dy}^2 = V_b^2 + (v_{rel} \sin \theta)^2$. Substituting this:
$$mgR(1 – \cos \theta) = \frac{1}{2}m [V_b^2 + v_{rel}^2 \sin^2 \theta] + \frac{1}{2}m V_b^2$$ $$gR(1 – \cos \theta) = V_b^2 + \frac{1}{2} v_{rel}^2 \sin^2 \theta$$Substitute $v_{rel} = \frac{2 V_b}{\cos \theta}$:
$$gR(1 – \cos \theta) = V_b^2 + \frac{1}{2} \left( \frac{4 V_b^2}{\cos^2 \theta} \right) \sin^2 \theta$$ $$gR(1 – \cos \theta) = V_b^2 \left( 1 + 2 \tan^2 \theta \right)$$We want to maximize $v_{dx} = V_b$. Let $u = \cos \theta$.
$$V_b^2 = \frac{gR(1 – u)}{1 + 2 \frac{1-u^2}{u^2}} = \frac{gR(1 – u) u^2}{u^2 + 2 – 2u^2} = \frac{gR(1 – u) u^2}{2 – u^2}$$3. Calculating the Critical Angle
To find the maximum, we differentiate $y = \frac{(1-u)u^2}{2-u^2}$ with respect to $u$ and set it to zero.
$$y = \frac{u^2 – u^3}{2 – u^2}$$ $$y’ = \frac{(2 – u^2)(2u – 3u^2) – (u^2 – u^3)(-2u)}{(2 – u^2)^2} = 0$$Numerator must be zero:
$$(2 – u^2)(2u – 3u^2) + 2u(u^2 – u^3) = 0$$ $$u [ (2 – u^2)(2 – 3u) + 2(u^2 – u^3) ] = 0$$Since $u = \cos \theta \neq 0$:
$$(4 – 6u – 2u^2 + 3u^3) + (2u^2 – 2u^3) = 0$$ $$u^3 – 6u + 4 = 0$$We need to solve $u^3 – 6u + 4 = 0$. By inspection, or checking the proposed answer $\cos \theta = \sqrt{3} – 1$:
Let $u = \sqrt{3} – 1$. Then $u^3 = (\sqrt{3}-1)^3 = 3\sqrt{3} – 9 + 3\sqrt{3} – 1 = 6\sqrt{3} – 10$.
Substitute into equation:
$$(6\sqrt{3} – 10) – 6(\sqrt{3} – 1) + 4$$ $$6\sqrt{3} – 10 – 6\sqrt{3} + 6 + 4 = 0$$The equation holds. Thus, the solution is:
$$\theta = \cos^{-1}(\sqrt{3} – 1)$$