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 ______
($g = 9.8\text{ m.s}^{-2}$)

Question

On a horizontal plane, a projectile of mass $m$ is launched from the ground with speed $v_0$ at an angle $\theta_0$ with the horizontal. In addition to the gravitational force ($mg$), it also experiences a drag force $\vec{F}_{drag} = -\gamma \vec{v}$, where $\vec{v}$ is its velocity and $\gamma$ is a constant. It hits the ground at a distance $R$ from the point of launch with its velocity making an angle $\theta$ with the horizontal, as shown schematically in the figure below.

Then which of the following options is correct?

Answer 1

Step 1: Get the spin rate at the moment of separation.
Before it leaves the table, the scale swings about the table edge like a rod pivoted at one end. Newton's law for rotation gives $\alpha = \tau/I$, and for a rod of length $L$ pivoted at an end, $I = mL^2/3$ while gravity supplies torque $\tau \approx mgL/2$ (valid since the turn angle stays small in this stage). This gives a constant angular acceleration:
\[ \alpha = \frac{3g}{2L} = \frac{3(9.8)}{2(0.15)} = 98\ \text{rad/s}^2 \]

Step 2: Use $\omega = \alpha t$ over the given 0.1 s window.
Starting from rest, after the stated $t_1 = 0.1$ s the rod has picked up a spin rate:
\[ \omega_0 = (98)(0.1) = 9.8\ \text{rad/s} \]
This is the value carried over into the next stage, since the pivot lets go right at this instant.

Step 3: Treat the second stage as independent free fall plus steady spin.
Once separated, the only force on the rod is gravity acting through its centre of mass, so there is zero net torque about the centre of mass and the spin rate $\omega_0$ stays fixed for the rest of the motion. The vertical drop of the centre of mass, since the rotation up to separation is negligible, can be treated on its own using simple free fall from rest:
\[ 0.5 = \frac{1}{2}(9.8)t_2^2 \implies t_2 = \sqrt{\frac{1}{9.8}} = 0.3194\ \text{s} \]

Step 4: Combine the two to get the angle turned.
Because the spin rate is constant through this stage, the angle swept equals the product of the spin rate and this time, independent of the vertical motion itself:
\[ \theta = \omega_0 t_2 = (9.8)(0.3194) = 3.130\ \text{rad} \]
Converting radians to degrees using $180/\pi$:
\[ \theta = 3.130\times 57.3^\circ = 179.4^\circ \]

Final Answer:
By the time the centre of mass has dropped 0.5 m, the scale has rotated by about $179.4^\circ$, which is nearly half a full turn. \[ \boxed{\theta = 179.4^{\circ}} \]

Show Hint

Correct Answer: 179.4

Solution and Explanation

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