A 15 cm long scale is held horizontally with one of its ends on the edge of a 1 m high table and the other end resting on one's index finger. As the finger is removed (see figure below), the scale starts rotating about its end on the table. After 0.1 s, during which it has rotated by a negligibly small angle but has gained a rotational speed, it leaves the table and falls vertically towards the ground. When its centre of mass has fallen by 0.5 m, it has rotated by an angle \(\theta\). The value of \(\theta\) in degrees (rounded off to one decimal place) is ______ (\(g = 9.8\text{ m.s}^{-2}\))
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Hint:
Find the angular speed gained while pivoted using \(\alpha = 3g/(2L)\) and \(\omega_0 = \alpha t_1\), then find the time to fall 0.5 m from rest, and multiply that time by \(\omega_0\) to get the angle turned.
Step 1: Understanding the Concept:
While the scale rests on the table edge with the finger removed, it behaves as a uniform rod pivoted at one end, falling under gravity. Since the problem tells us the angle turned in the first 0.1 s is negligibly small, we can treat the torque during this stage as constant, using the horizontal orientation throughout.
Step 2: Find the angular speed gained while pivoted.
For a uniform rod of length \(L\) pivoted at one end, the moment of inertia about that end is \(I = \frac{1}{3}mL^2\). Gravity acts at the centre of mass, a distance \(L/2\) from the pivot, giving torque \(\tau = mg(L/2)\) (using \(\cos\theta \approx 1\) since the angle stays small). So the angular acceleration is:
\[ \alpha = \frac{\tau}{I} = \frac{mg(L/2)}{\frac{1}{3}mL^2} = \frac{3g}{2L} \]
With \(g=9.8\text{ m/s}^2\) and \(L=0.15\text{ m}\):
\[ \alpha = \frac{3\times 9.8}{2\times 0.15} = 98\text{ rad/s}^2 \]
Starting from rest, after \(t_1 = 0.1\) s the scale has gained an angular speed:
\[ \omega_0 = \alpha t_1 = 98\times 0.1 = 9.8\text{ rad/s} \]
Step 3: Track the free-fall stage.
Once it leaves the table, no torque acts on the scale about its own centre of mass (gravity acts right at the centre of mass), so the angular speed stays fixed at \(\omega_0 = 9.8\) rad/s for the rest of the fall. Since the rotation is negligible at the moment of separation, the centre of mass effectively begins this stage of falling from rest, and drops under gravity alone.
Step 4: Find the time to fall 0.5 m and the angle turned.
Using \(h = \frac{1}{2}gt_2^2\) with \(h = 0.5\) m:
\[ t_2 = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2\times 0.5}{9.8}} = \sqrt{0.1020} = 0.319\text{ s} \]
The scale keeps spinning at the constant rate \(\omega_0\) throughout this fall, so the angle turned is:
\[ \theta = \omega_0 t_2 = 9.8\times 0.319 = 3.130\text{ rad} \]
Convert to degrees:
\[ \theta = 3.130\times\frac{180}{\pi} = 179.4^{\circ} \]
Final Answer:
The scale has turned by about \(179.4^{\circ}\) by the time its centre of mass has fallen 0.5 m.
\[ \boxed{\theta = 179.4^{\circ}} \]
