A rod of length l = 1 m rotates with constant angular…

A rod of length $l = 1\ \mathrm{m}$ rotates with constant angular velocity $\omega = 314\ \mathrm{rad/s}$ about an axis through one end. A small object comes off the other end and flies off along the tangent, at $\alpha = 45^\circ$ to the horizontal. (a) How far away does it land on the horizontal plane? (b) How high does it rise? (Treat the launch point as being on the plane; no air resistance.)

Explanation of Method

An object flung from a rotating rod.

Step 1: Launch speed

The end of the rod moves on a circle with speed $v = \omega l$, tangent to the circle: $v = 314 \cdot 1 = 314\ \mathrm{m/s}$.

\[ v_x = v_y = 314 \cdot \tfrac{\sqrt2}{2} \approx 222.0\ \mathrm{m/s}. \]

Step 2: (a) Range

\[ d = \frac{v^2\sin 90^\circ}{g} = \frac{314^2}{9.81} = \frac{98\,596}{9.81}. \]

\[d \approx 10\,050\ \mathrm{m} \approx 10.05\ \mathrm{km}\]

Step 3: (b) Maximum height

\[ H = \frac{v^2\sin^2 45^\circ}{2g} = \frac{v^2}{4g} = \frac{98\,596}{39.24}. \]

\[H \approx 2513\ \mathrm{m}\]

$l = 1\ \mathrm{m}$ $v = \omega l = 314\ \mathrm{m/s},\ 45^\circ$ $\omega$ $\approx 10.05\ \mathrm{km}$ $2.51\ \mathrm{km}$

Left: the object leaves the end of the rod along the tangent with the tip speed $\omega l$. Right: its free flight, drawn to scale (the rod's size is negligible on this scale).

Result

\[\boxed{\,d \approx 10\,050\ \mathrm{m} \approx 10.05\ \mathrm{km}\,}\]

\[\boxed{\,H \approx 2513\ \mathrm{m}\,}\]

Remark

Note. The flight takes $2v_y/g \approx 45.3\ \mathrm{s}$. The original values (10 039 m and 2510 m) differ slightly because $314 \cdot 0.7071$ was rounded to 221.94 instead of 222.03. At 45° the height is always a quarter of the range.