A meter stick held vertically, with one end on the ground, is then allowed to fall. What is the value of the radial and tangential acceleration of the top end of the stick when the stick has turned through on angle $\theta?$ What is the speed with which the top end of the stick hits the ground? Assume that the end of the stick in. contact with the ground does not slip.
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Let OA be the initial vertical position of the stick of length I with the end O in contact with ground. As the stick falls, it revolves around point O. OB represents the position of the stick at any time t = t when it has turned thorough an angle $\theta.$
C and C’ are the centre of mass of the stick at t = 0 and t = t respectively. The torque of the weight mg of the stick at t = t about O is
$=\text{mg}\times\text{x}=\text{mg}\frac{\text{l}}{2}\sin\theta\dots(1)$
If I and α denote the moment of inertia, and angular acceleration of stick, we have
$\tau=\text{I}\alpha\text{ or }\alpha=\frac{\tau}{\text{I}}=\frac{\text{mgl}}{2\text{}}\sin\theta\dots(2)$
The moment of inertia, I of stick about axis passing through one end O is
$\text{I}=\frac{\text{ml}^2}{3}$
$\therefore\alpha=\frac{3}{2}\frac{\text{g}}{\text{l}}\sin\theta\dots(3)$
Now, by definition $\alpha=\frac{\text{d}\omega}{\text{dt}}=\frac{\text{d}\omega}{\text{d}\theta}.\frac{\text{d}\theta}{\text{dt}}=\omega\frac{\text{d}\omega}{\text{d}\theta}\dots{4}$
From eqns. (3) and (4) $\omega\text{ a }\omega=\frac{3}{2}\frac{\text{g}}{\text{l}}\sin\theta\text{ d }\theta\dots(5)$
Intergrating, we get $\frac{\omega^2}{2}=-\frac{3}{2}\frac{\text{g}}{\text{l}}\cos\theta+\text{C}_1\dots(6)$
where $C_1$ is a constant of integration. Its value is found from the initial condition i.e., at $\text{t}=0;\theta=0\text{ and }\omega=0$
$\therefore0=-\frac{3}{2}\frac{\text{g}}{\text{l}}+\text{C}_1\text{ or }\text{C}_1=\frac{3}{2}\frac{\text{g}}{\text{l}}\dots(7)$
Combining Eqns. (6) and (7), we get
$\omega^2=\frac{3\text{g}}{\text{l}}(1-\cos\theta)\dots(8)$
The radial acceleration $a_r$ is
$\text{a}_{\text{r}}=\omega^2\text{l}=3\text{g}(1-\cos\theta)$
The tangential acceleration $a_1$ is
$\text{a}_{\text{t}}=\text{l}\alpha=\frac{3}{2}\text{g}\sin\theta\dots(10)$
Iet $\omega_0$ be the angular speed of the end A of stick as it this the ground. (The stick has turned through $\theta=90^\circ).$ Then
$\omega_0=3\text{g}\dots(11)$
The linear speed $v_0$ of the end A of stick as it hits are ground is $\text{v}_0=\text{l}\omega_0=3\text{gl}\dots(12)$
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