Mechanics + Magnetism = Mechatism 2

Diagram

As Shown in figure 'O' is ICR ( Instantaneous centre of rotation ) About This Point motion of rod is Pure Rotatory .

So for calculating Torque about 'O' is only due to $mg\sin { \theta }$ . component since all other components Do not create Torque.

${ \tau }_{ 0 }\quad =\quad { I }_{ 0 }\quad \alpha \\ mg\sin { \theta } \frac { L }{ 2 } =\quad (\frac { m{ L }^{ 2 } }{ 12 } \quad +\quad \frac { m{ L }^{ 2 } }{ 4 } )\alpha \\ \\ \alpha \quad =\quad \frac { 3g\sin { \theta } }{ 2L } \\ \\ \quad -\omega \frac { d\omega }{ d\theta } \quad =\quad \frac { 3g\sin { \theta } }{ 2L } \\ \quad \quad \quad \\ \int _{ 0 }^{ \omega }{ \omega d\omega } \quad =\quad -\quad \frac { 3g }{ 2L } \int _{ { 60 }^{ 0 } }^{ { 30 }^{ 0 } }{ \sin { \theta } d } \theta \\ \\ \omega \quad =\sqrt { \frac { 3g }{ 2L } (\sqrt { 3 } -1) } \quad \quad \quad \longrightarrow \quad (1)$ .

NOTE

( You Can Also Use Energy conservation For Calculating angular velocity )

Now For Calculating EMF Between A & P We Calculate emf 's between OA and OP & then Subtract them to get emf between A & P since They are connected in Series with opposite Polarity

${ E }_{ OA }\quad =\quad \frac { B\omega { ({ l }_{ OA }) }^{ 2 } }{ 2 } \\ \\ \quad \quad \quad \quad =\quad \frac { B\omega { (L\cos { { (30 }^{ 0 }) } ) }^{ 2 } }{ 2 } \\ \quad \quad \quad \quad \\ \quad \quad \quad \quad =\quad \frac { 3B\omega { (L) }^{ 2 } }{ 8 } \\ \\ { E }_{ OP }\quad =\quad \frac { B\omega { ({ { l }_{ op } }) }^{ 2 } }{ 2 } \\ \quad \quad \quad \quad =\quad \frac { 7B\omega { (L) }^{ 2 } }{ 32 } \\ \\ { E }_{ AP }\quad =\quad { E }_{ OA }\quad -\quad { E }_{ OP }\\ \quad \quad { E }_{ AP }\quad \quad =\quad \frac { 5B\omega { (L) }^{ 2 } }{ 32 } \quad \longrightarrow (2)$ .

For calculation Of OP see below

Now Putting all value in 2 equation we get

$\\ { E }_{ AP }\quad =\quad \frac { 5B{ (L) }^{ 2 } }{ 32 } \sqrt { \frac { 3g }{ 2L } (\sqrt { 3 } -1) } \\ \quad \quad \quad \quad =\quad 0.517714\\ \\ \quad \left\lfloor 100\alpha \right\rfloor \quad =\quad 51\quad Ans.$ .

Note : For Calculation Length of OP we use Cosine Law in triangle OCP at the instant of time when angle is 30 degree from horizontal as shown in figure below.

Fig

$\cos { { 120 }^{ 0 } } \quad =\quad \frac { { (\frac { L }{ 2 } ) }^{ 2 }+{ (\frac { L }{ 4 } ) }^{ 2 }-{ (OP) }^{ 2 } }{ 2(\frac { L }{ 2 } )(\frac { L }{ 4 } ) } \\ \Rightarrow \quad \quad { OP }^{ 2 }\quad =\quad \frac { { 7L }^{ 2 } }{ 16 } \quad \\ \\$ .

Instead of a trick with ICR of the rod I'll use a lagrangian to derive equations of motion.

Let $A=(0,y)$ and $B=(x,0)$ be coordinates of points $A$ and $B$ . We easily observe that $x^2+y^2=l$ . A point $P$ on a rod such that $|AP|=\lambda l$ ( $0\leq \lambda \leq 1$ ) has coordinates $P=(\lambda x,(1-\lambda y))$ . Therefore, its velocity vector $\vec{v}(\lambda)$ has coordinates $\vec{v}(\lambda)=(\lambda \dot{x}, (1-\lambda)\dot{y})$ . So the kinetic energy $T$ has value $T=\frac{m}{2}\int _0^1\left(\lambda^2\dot{x}^2+(1-\lambda)^2\dot{y}^2\right)d\lambda=\frac{m}{6}(\dot{x}^2+\dot{y}^2)$ . Now, we take $x=l\cos\theta$ and $y=l\sin\theta$ . This gives us $\dot{x}=-l\dot{theta}\sin\theta$ and $\dot{y}=l\dot{\theta}\cos\theta$ . So we have $T=\frac{m}{6}l^2\dot{\theta}^2$ . Potential energy $V$ of rod is equal to $V=\frac{1}{2}mgl\sin\theta$ . So the lagrangian $\mathcal{L}$ is $\mathcal{L}=\frac{m}{6}l^2\dot{\theta}^2-\frac{1}{2}mgl\sin\theta$ This gives us equation of motion $\frac{m}{3}l^2\ddot{\theta}=-\frac{1}{2}mgl\cos\theta$ . Now we use $\ddot{\theta}=\dot{\theta}\frac{d\dot{\theta}}{d\theta}$ which yields $\frac{m}{3}l^2\int _{0}^{\omega}\dot{\theta}d\dot{\theta}=-\frac{1}{2}mgl\int _{\frac{\pi}{3}}^{\frac{\pi}{6}}\cos\theta d\theta$ . After this computation we get $\omega=\sqrt{\frac{3g}{2l}(\sqrt{3}-1)}$ .

Since each point on the rod has velocity $\vec{v}(\lambda)=(-\lambda l\dot{\theta}\sin\theta, (1-\lambda)l\dot{\theta}\cos\theta)$ its velocity $v_{\perp}(\lambda)$ is equal to $v_{\perp}(\lambda)=|-\lambda l\dot{\theta}\sin^2\theta +(1-\lambda)l\dot{\theta}\cos^2\theta)|$ . Now it suffice to take $d\mathcal{E}=Blv_{\perp}(\lambda)$ , and substitute $\theta=\frac{\pi}{6}$ , $\dot{\theta}=\omega$ and calculate $\mathcal{E}_{AP}=Bl\int_{0}^{\frac{1}{4}}v_{\perp}(\lambda)d\lambda=\frac{5}{32}Bl^2\sqrt{\frac{3g}{2l}(\sqrt{3}-1)}$ .