Rotational Motion

The relative velocity of the point $P$ w.r.t. the point $Q$ is given by

$ \vec{v}_r=\vec{v}_P-\vec{v}_Q $ .........(1)

It is easy to see that $\left|\vec{v}_P\right|=\left|\vec{v}_Q\right|=\omega R$ and angle traversed in time $t$ is $\omega t$. Thus, velocities of $P$ and $Q$ are

$ \begin{aligned} & \vec{v}_P=\omega R(-\sin \omega t \hat{\imath}-\cos \omega t \hat{\jmath}) \\ & \vec{v}_Q=\omega R(\sin \omega t \hat{\imath}-\cos \omega t \hat{\jmath}) \end{aligned} $

Substitute $\vec{v}_P$ and $\vec{v}_Q$ in equation (1) to get $\vec{v}_r=$ $-2 \omega R \sin \omega t \hat{\imath}$ and thus $v_r=2 \omega R|\sin \omega t|$.

$\left(\frac{1}{8}\right)^{\mathrm{th}}$ rotation of the disc with a constant $\Omega$ implies that the disc has turned through

$ \frac{1}{8} \times 360^{\circ}=45^{\circ} $

The orientation of the disc now is

For the particle thrown from $Q$ towards $R$, will have only the velocity given to it with respect to disc i.e., it possess a velocity only in the Y-Z plane and covers a range along OR. Hence, it can be seen to land on the disc in the unshaded region. Actually point $O$ of the disc has zero velocity and $Q$ being very close to $O$, will have only the velocity given to it w.r.t disc.

For the particle thrown from P , will have an additional velocity $=R \Omega$ along +X direction i.e., it will cover a range $=\frac{R}{2}$ along PO as well as a distance $=(R \omega) \frac{T}{8}=\frac{R}{8} \times 2 \pi=\frac{\pi R}{4}$ along $+X$ direction

The resultant displacement of particle P is thus obtained as

$\begin{aligned} \Rightarrow \quad d & =\frac{R}{2} \sqrt{1+\frac{\pi^2}{4}} \\ \Rightarrow \quad \pi^2 & =10 \\ d & =\frac{\mathrm{R}}{4} \sqrt{14} \\ \tan \theta & =\frac{\pi \mathrm{R} / 4}{\mathrm{R} / 2}=\frac{\pi}{2}\end{aligned}$

In the diagram shown,

$ \begin{aligned} P S & =R \cos \theta \\ & =R \times \frac{2}{\sqrt{14}} \\ & =\frac{2 R}{\sqrt{14}} \end{aligned} $

Since, $d>$ PS $\quad$ PS $=\frac{R}{7} \sqrt{14}$

Hence, the particle from P also lands in the unshaded region.

As depicted in the figure shown here, when the system, as a whole, turns by 180$^\circ$, the disc also turns by 180$^\circ$ about its vertical axis. Hence, the instantaneous axis that passes through the centre of mass is vertical in both cases (a) and (b).

As depicted in the figure shown here, when the system, as a whole, turns by 180$^\circ$, the disc also turns by 180$^\circ$ about vertical axis. Hence, for both cases (a) and (b), the angular speed about the instantaneous axis that passes through the centre of mass is $\omega$.

Angular momentum of a particle about a point is given by:

L = r $\times$ p = m (r $\times$ v)

For L_O

$\left| L \right| = (mvr\sin \theta ) = m(R\omega )(R)\sin 90^\circ $

$ = m{R^2}\omega $ = constant

Direction of L_O is always upwards. Therefore, complete L_O is constant, both in magnitude as well as direction.

For L_P

$\left| {{L_P}} \right| = (mvr\sin \theta ) = m(R\omega )(l)\sin 90^\circ $

$ = (mRl\omega )$

Magnitude of L_P will remain constant but direction of L_P keeps on changing.

Let M and l be the mass and length of the rod respectively and m be the mass of the insect. Let the insect be at a distance x from O at any instant of time t.

$\therefore$ x = vt ....... (i)

Angular momentum of the system about O,

$L = \left[ {{{M{l^2}} \over {12}} + m{x^2}} \right]\omega $

$ = \left[ {{{M{l^2}} \over {12}} + m{{(vt)}^2}} \right]\omega $ (Using (i))

$ = \left[ {{{M{l^2}} \over {12}} + m{v^2}{t^2}} \right]\omega $

As, $\left| {\overrightarrow \tau } \right| = {{dL} \over {dt}} = {d \over {dt}}\left[ {{{M{l^2}} \over {12}} + m{v^2}{t^2}} \right]\omega $

As $\omega$ and v remain constant

$\therefore$ $\left| {\overrightarrow \tau } \right| = 2m\omega {v^2}t$

$\left| {\overrightarrow \tau } \right| \propto t$

Hence, the graph and t is a straight line passing through the (0, 0). Option (b) represents correct plot.

The outer ring rolls without slipping, meaning the point on the ring in contact with the ground (denoted as point C) is stationary, i.e., $ \vec{v}_{\mathrm{C}} = \overrightarrow{0} $. Therefore, the velocity of point O is :

$ \vec{v}_{\mathrm{O}} = \vec{v}_{\mathrm{C}} + \vec{\omega}_{\mathrm{o}} \times \vec{r}_{\mathrm{CO}} = \overrightarrow{0} + \omega \hat{\jmath} \times 3R \hat{k} = 3R\omega \hat{\imath} $

Point P is located on the inner disc, which has an angular velocity $ \vec{\omega}_{\mathrm{i}} = -\omega / 2 \hat{\jmath} $. The position vector from O to P is given by :

$ \vec{r}_{\mathrm{OP}} = R \cos 30^{\circ} \hat{\imath} + R \sin 30^{\circ} \hat{k} = \frac{\sqrt{3} R}{2} \hat{\imath} + \frac{R}{2} \hat{k} $

Thus, the velocity of point P is calculated as follows :

$ \begin{aligned} \vec{v}_{\mathrm{P}} & = \vec{v}_{\mathrm{O}} + \vec{\omega}_{\mathrm{i}} \times \vec{r}_{\mathrm{OP}} \\ & = 3R\omega \hat{\imath} + \left(-\frac{\omega}{2} \hat{\jmath}\right) \times \left(\frac{\sqrt{3} R}{2} \hat{\imath} + \frac{R}{2} \hat{k}\right) \\ & = \frac{11\omega R}{4} \hat{\imath} + \frac{\sqrt{3} \omega R}{4} \hat{k} . \end{aligned} $

Since Cylinder P has most of its mass concentrated near its surface, its moment of inertia (about the cylinder axis) is greater than that of Cylinder Q, i.e., $ I_P > I_Q $. The forces acting on the cylinder include its weight $ mg $, the normal reaction $ N $, and the frictional force $ f $.

In the case of rolling without slipping, we have :

$ v = \omega r $

$ a = \alpha r $

The torque about the center of mass is related to $\alpha$ by :

$ \tau = rf = I\alpha \quad (1) $

The force along the plane is related to $a$ by :

$ mg \sin \theta - f = ma \quad (2) $

By solving equations (1) and (2), we get :

$ a = \frac{g \sin \theta}{1 + \frac{I}{mr^2}} \quad (3) $

From equation (3), it follows that $ a_P < a_Q $ (since $ I_P > I_Q $). Therefore, Cylinder P reaches the ground later, has a lower velocity, lower angular velocity ($\omega = \frac{v}{r}$), and lower translational kinetic energy.

For verification, the conservation of energy can be used :

$ mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 $

This yields :

$ \frac{1}{2}mv_P^2 < \frac{1}{2}mv_Q^2 $

Thus :

$ v_P < v_Q, \ \omega_P < \omega_Q, \ a_P < a_Q $

And the time relationships :

$ t_P > t_Q $

We have

Restoring torque = Force of gravity on disc and rod which is same in both cases.

$\bullet$ Case A : The moment of inertia is

${I_A} = {{M{R^2}} \over 2} + M{L^2} + {{m{l^3}} \over 3}$

Therefore,

${\tau _A} = - {I_A}\omega _A^2\theta = - \left( {{{M{R^2}} \over 2} + M{L^2} + {{m{l^3}} \over 3}} \right)\omega _A^2\theta $

$\bullet$ Case B : The moment of inertia is

${I_B} = {{m{l^3}} \over 3} + M{L^2}$

Therefore,

${\tau _B} = - {I_B}\omega _B^2\theta = - \left( {{{m{l^3}} \over 3} + M{L^2}} \right)\omega _B^2\theta $

since ${\tau _A} = {\tau _B} \Rightarrow {\omega _A} < {\omega _B}$.

We have

$\overrightarrow {{V_C}} = \overrightarrow {{V_{CM}}} + \overrightarrow {r\omega } = 2\overrightarrow {{V_{CM}}} $

$\overrightarrow {{V_B}} = \overrightarrow {{V_{CM}}} $

$\overrightarrow {{V_A}} = 0$

Therefore,

$\overrightarrow {{V_C}} - \overrightarrow {{V_B}} = \overrightarrow {r\omega } = \overrightarrow {{V_{CM}}} $

$\overrightarrow {{V_B}} - \overrightarrow {{V_A}} = \overrightarrow {{V_{CM}}} $

$\overrightarrow {{V_C}} - \overrightarrow {{V_A}} = \overrightarrow {{V_{CM}}} + \overrightarrow {r\omega } = 2\overrightarrow {{V_{CM}}} $

$\overrightarrow {{V_B}} - \overrightarrow {{V_C}} = - \overrightarrow {r\omega } $

For pure rolling, we have

$|\overrightarrow {{V_{CM}}} | = |\overrightarrow {r\omega } |$

Therefore,

$|\overrightarrow {{V_C}} - \overrightarrow {{V_A}} | = 2{V_{CM}}$

$ \Rightarrow 2|\overrightarrow {{V_B}} - \overrightarrow {{V_C}} | = 2r\omega = 2{V_{CM}}$

According to Newton's second law, we have

$\sum {{{\overrightarrow f }_{ext}} = {{d\overrightarrow p } \over {dt}}} $

If $\sum {{{\overrightarrow f }_{ext}} = 0} $, we find that the linear momentum $\overrightarrow p $ as constant.

When the disc is at a distance x from the mean position (equilibrium position), the force acting on the disc are give below.

$\therefore -2kx+f=-M_{ac}$ ..... (i)

Where $a_c$ = acceleration of centre of mass.

The torque acting on the disc about its centre of mass C is

$\tau = f \times R = I \times {a_c}$

$\therefore$ $f = {{I\alpha } \over R} = {{{1 \over 2}M{R^2}} \over R} \times {{{a_c}} \over R}$

[$\therefore$ $I = {1 \over 2}M{R^2},{a_c} = R{a_c}$ for rolling without slipping]

$\therefore$ $f = {1 \over 2}M{a_c}$ ...... (ii)

From eq. (i) and (ii),

$ - 2kx + {1 \over 2}M{a_c} = - M{a_c}$

$ \Rightarrow {3 \over 2}M{a_c} = 2kx$

$ \Rightarrow M{a_c} = {{4kx} \over 3}$

$\Rightarrow$ Net external force acting on the disc when its centre of mass is at displacement x with respect to the equilibrium position $ = {{ - 4kx} \over 3}$ directed towards the equilibrium.

When the disc is at a distance x from the mean position (equilibrium position), the force acting on the disc are give below.

$\therefore -2kx+f=-M_{ac}$ ..... (i)

Where $a_c$ = acceleration of centre of mass.

The torque acting on the disc about its centre of mass C is

$\tau = f \times R = I \times {a_c}$

$\therefore$ $f = {{I\alpha } \over R} = {{{1 \over 2}M{R^2}} \over R} \times {{{a_c}} \over R}$

[$\therefore$ $I = {1 \over 2}M{R^2},{a_c} = R{a_c}$ for rolling without slipping]

$\therefore$ $f = {1 \over 2}M{a_c}$ ...... (ii)

From eq. (i) and (ii),

$ - 2kx + {1 \over 2}M{a_c} = - M{a_c}$

$ \Rightarrow {3 \over 2}M{a_c} = 2kx$

$ \Rightarrow M{a_c} = {{4kx} \over 3}$

$\Rightarrow$ Net external force acting on the disc when its centre of mass is at displacement x with respect to the equilibrium position $ = {{ - 4kx} \over 3}$ directed towards the equilibrium.

$ \therefore $ $|{F_{net}}| = {{4k} \over 3}x$

For S.H.M $|{F_{net}}| = M{\omega ^2}x$

$\therefore$ $M{\omega ^2} = {{4k} \over 3}$

$ \Rightarrow \omega = \sqrt {{{4k} \over {3M}}} $ .... (iii)

When the disc is at a distance x from the mean position (equilibrium position), the force acting on the disc are give below.

$\therefore -2kx+f=-M_{ac}$ ..... (i)

Where $a_c$ = acceleration of centre of mass.

The torque acting on the disc about its centre of mass C is

$\tau = f \times R = I \times {a_c}$

$\therefore$ $f = {{I\alpha } \over R} = {{{1 \over 2}M{R^2}} \over R} \times {{{a_c}} \over R}$

[$\therefore$ $I = {1 \over 2}M{R^2},{a_c} = R{a_c}$ for rolling without slipping]

$\therefore$ $f = {1 \over 2}M{a_c}$ ...... (ii)

From eq. (i) and (ii),

$ \Rightarrow - 2kx + f = - 2f$

$ \Rightarrow f = {{2k} \over 3} \times x$

Frictional force depends on x. As x increase friction increase. Frictional force (f) is maximum at $x = A$.

Where A = amplitude of S.H.M

Maximum frictional force,

${f_{\max }} = {{2k} \over 3} \times A$

The force should be almost equal to the limiting friction ($\mu mg$) for rolling without slipping.

$\therefore$ $\mu mg = {{2k} \over 3} \times A$ ..... (iv)

For S.H.M

Velocity amplitude $ = A\omega $

$\therefore$ ${V_0} = A\omega $

$\therefore$ ${V_0} = {{3m\,mg} \over {2k}}w$ (from (iv))

$\therefore$ ${V_0} = {{3\mu mg} \over {2k}} \times \sqrt {{{4k} \over {3M}}} $ (As $ \omega = \sqrt {{{4k} \over {3M}}} $)

$\therefore$ ${V_0} = mg\sqrt {{{3M} \over k}} $

The acceleration of a body rolling down an inclined plane in given by

$a = {{g\sin \theta } \over {1 + {I \over {M{R^2}}}}}$

For hollow cylinder,

${I \over {M{R^2}}} = {{M{R^2}} \over {M{R^2}}} = 1$

For solid cylinder,

${1 \over {M{R^2}}} = {{{1 \over 2}M{R^2}} \over {M{R^2}}} = {1 \over 2}$

$\therefore$ Acceleration of the solid cylinder is more than that of hollow cylinder and therefore solid cylinder will reach the bottom of the inclined plane first.

$\therefore$ Statement - 1 is False.

Statement - 2 : In case of rolling, there will be no heat losses. Therefore, total mechanical energy remains conserved. The potential energy, therefore gets converted into kinetic energy. In both the cases, since the initial potential energy is same, the final kinetic energy will also be same. Therefore, statement - 2 is correct.

By applying conservation of Energy, we have,

$\frac{1}{2} m v^{2}+\frac{1}{2} \mathrm{I} \omega^{2}=m g h_{\max }$ ..... (i)

For pure rolling,

$v=R\omega$

$\Rightarrow \quad \omega=v / \mathrm{R}$ ..... (ii)

Putting eq. (ii) in eq. (i), we get

$\Rightarrow \frac{1}{2} m v^{2}+\frac{1}{2} \mathrm{I}\left(\frac{v^{2}}{R^{2}}\right)=m g \times \frac{3 v^{2}}{4 g}$ $\left(\because h_{\max } \times \frac{3 v^{2}}{4 g}\right)$

$\Rightarrow \quad \mathrm{I}=\frac{1}{2} \mathrm{MR}^{2}$

Hence, this is a required formula of moment of Inertia of Disc. Therefore, the object a disc.

$2^{\text {nd }}$ Method : For pure rolling, the velocity of point of contact is zero i.e., $v=0$, So, $\Delta x$ (displacement) is zero. Then work done by the frictional force should be

$W=f_{0} \Delta x=0$

If work is done by dissipative force, work done should be zero. It means that work done of system is zero. So energy should be conserved. Initially, the object has both translation as well as rotational motion. If it attains '$h_{\max }$' kinetic energy changes into potential energy.

From energy conservation principle,

$\mathrm{K} . \mathrm{E}_{\text {translational }}+\mathrm{K} . \mathrm{E}_{\text {rotational }}=$ Potential Energy

$\Rightarrow \quad \frac{1}{2} m v^{2}+\frac{1}{2} \mathrm{I} \omega^{2}=m g \times \frac{3 v^{2}}{4 g}$

$\Rightarrow \quad m v^{2}+\mathrm{I} \omega_{2}=\frac{3 v^{2}}{2} m$

$\Rightarrow \quad \mathrm{I} \omega^{2}=\frac{1}{2} m v^{2}$

$\Rightarrow \quad \mathrm{I}=\frac{1}{2} \frac{m v^{2}}{\omega^{2}}$ ..... (i)

For No slipping,

$\Rightarrow$ $\begin{aligned} & v=\omega R \\ & \omega=\frac{v}{R}\end{aligned}$ ..... (ii)

Put eq (ii) in eq (i)

Hence, $\quad \mathrm{I}=\frac{1}{2} \mathrm{MR}^{2}$

$\mathrm{I}=\frac{1}{2} \mathrm{M} \frac{v^{2}}{v^{2}} \times \mathrm{R}^{2}$

this is moment of inertia of disc.

So, object is a Disc.

Statement 1 : For velocity of centre of mass remains constant, net force acting on body must be zero. Therefore the statement 1 is false.

Statement 2 : The linear momentum of an isolated system remains constant. The statement is true.

For disc A,

${1 \over 2}kx_1^2 = {1 \over 2}I{(2\omega )^2}$

$ \Rightarrow kx_1^2 = 2I{\omega ^2}$ ..... (i)

For disc B,

${1 \over 2}kx_2^2 = {1 \over 2}2I{\omega ^2}$

$ \Rightarrow kx_2^2 = I{\omega ^2}$ ..... (ii)

On dividing eq. (i) by (ii)

${{kx_1^2} \over {kx_2^2}} = {{2I{\omega ^2}} \over {I{\omega ^2}}}$

${{{x_1}} \over {{x_2}}} = \sqrt 2 $

When disc B is brought in contact with disc A. Let w' be the final angular velocity of both the disc rotating together.

Applying conservation of angular momentum for the two disc system.

$ \Rightarrow I(2\omega ) + 2I(\omega ) = (I + 2I)\omega '$

$ \Rightarrow \omega ' = {4 \over 3}\omega $

Torque on disc A,

${T_A} = {{\Delta {L_A}} \over t} = {{{L_f} - {L_i}} \over t} = {{I \times {4 \over 3}\omega - I \times 2\omega } \over t}$

$ = {{ - 2I\omega } \over {3t}}$

Note : The negative sign represents that the torque creates angular retardation.

Loss in kinetic energy

= (K.E.)$_{initial}$ $-$ (K.E.)$_{final}$

$ = \left[ {{1 \over 2}I{{(2\omega )}^2} + {1 \over 2}(2I){\omega ^2}} \right] - \left[ {{1 \over 2}(I + 2I){{\left( {{4 \over 3}\omega } \right)}^2}} \right]$

$ = [2I{\omega ^2} + I{\omega ^2}] - \left[ {{3 \over 2}I \times {4 \over 3} \times {4 \over 3}{\omega ^2}} \right]$

$ = 3I{\omega ^2} - {8 \over 3}I{\omega ^2} = {{I{\omega ^2}} \over 3}$

${I_{A}} = 2 \times m{\left( {{\textstyle{\ell \over {\sqrt 2 }}}} \right)^2} + m{\left( {\sqrt 2 \ell } \right)^2}$

$ = m{\ell ^2} + 2m{\ell ^2} = 3m{\ell ^2}$

To solve this problem, we need to compare the moments of inertia of the sphere and the disc. For a solid sphere of radius $R$, the moment of inertia about its geometrical axis is given by:

$I_{\text{sphere}} = \frac{2}{5} M R^2$

where $M$ is the mass of the sphere.

When the sphere is melted into a disc of radius $r$ and thickness $t$, the volume (and hence mass) remains the same. So, the volume of the sphere is:

$\frac{4}{3} \pi R^3$

And the volume of the disc is:

$\pi r^2 t$

Equating the volumes, we get:

$\frac{4}{3} \pi R^3 = \pi r^2 t$

This simplifies to:

$t = \frac{4 R^3}{3 r^2}$

Next, we need to find the moment of inertia of the disc about the tangential axis (which is perpendicular to the plane of the disc). The moment of inertia of a disc of mass $M$ and radius $r$ about an axis through its center and perpendicular to the plane of the disc is:

$I_{\text{disc, center}} = \frac{1}{2} M r^2$

Using the parallel axis theorem, the moment of inertia about a tangential axis is:

$I_{\text{disc, tangential}} = I_{\text{disc, center}} + M r^2 = \frac{1}{2} M r^2 + M r^2 = \frac{3}{2} M r^2$

Since we are given that the moment of inertia of the sphere and the disc about their respective axes are equal, we set:

$\frac{2}{5} M R^2 = \frac{3}{2} M r^2$

Canceling the mass $M$ from both sides and solving for $r$, we get:

$\frac{2}{5} R^2 = \frac{3}{2} r^2$

Solving for $r$, we have:

$r^2 = \frac{4}{15} R^2$

Taking the square root of both sides:

$r = \frac{2}{\sqrt{15}} R$

So, the correct answer is:

Option A: $\frac{2}{\sqrt{15}} R$

$ \text { From the diagram, } $

We observe $f$ opposes translational motion and (option d) supports rotational motion.

$ \begin{aligned} m g \sin \theta-f & =m a \\ f \mathrm{R} & =\mathrm{I} \alpha=\frac{1}{2} \mathrm{MR}^2\left(\frac{a}{\mathrm{R}}\right)=\frac{1}{2} m a \mathrm{R} \end{aligned} $

$ \begin{array}{rlrl} m g \sin \theta-f & =2 f & (m a & =2 f) \\ m g \sin \theta & =3 f & {[a} & =r \alpha \\ \alpha & \left.=\frac{a}{r}\right] & \end{array} $

$ \begin{aligned} &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, f=\frac{1}{3} m g \sin \theta \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, f \propto \sin \theta \end{aligned} $

Thus, if we reduce $\theta, f$ will be decreased Option (c) is correct.

A bullet of mass $m$ strikes the wooden log of mass M and length L and sticks to it.

Angular momentum of the system, about point O, remains conserved. Initial angular momentum = Final angular momentum.

$\therefore$ $mv \times L = {I_0}\omega $ ..... (i)

${I_0} = {I_{\log }} + {I_b}$

${I_0} = {{M{L^2}} \over 3} + m{L^2}$

${I_0} = {L^2}\left( {{M \over 3} + m} \right) = {L^2}\left( {{{M + 3m} \over 3}} \right)$

$\therefore$ $mvL = {L^2}\left( {{{M + 3m} \over 3}} \right)\omega $

$\omega = {{3mv} \over {L(M + 3m)}}$

Acceleration of a body rolling down the inclined plane

$a = {{g\sin \theta } \over {1 + {I \over {m{R^2}}}}}$

For cylinder, $\quad \mathrm{I}=\frac{1}{2} m \mathrm{R}^{2}$

$a = {{g\sin \theta } \over {1 + {{{1 \over 2}m{R^2}} \over {m{R^2}}}}} = {{g\sin \theta } \over {1 + {1 \over 2}}}$

$a = {{g\sin \theta } \over {{3 \over 2}}}$

$a = {2 \over 2}g\sin \theta $

Let the block of mass $m$ hangs from the point $P$ by a string attached to the hinges of the two ladders.

Newton's second law gives the tension $T$ in the string as $T=m g$. By symmetry, string tension pulls down each ladder by a force

$ T / 2=m g / 2 . $

By Newton's third law, the reaction forces acting on the two ladders at the hinge point $P$ are equal and opposite. These are shown by $R_1$ and $R_2$ in the figure. By symmetry, the normal reaction $N$ at $A$ and $B$ are equal, the friction forces $f$ at $A$ and $B$ are equal in magnitude but opposite in direction. Another force acting on both the ladders is their weight $M g$ which pass through their centre of mass. In equilibrium, the resultant forces on the two ladders are zero i.e.,

$ \begin{aligned} & N+R_2-M g-m g / 2=0, ........(1)\\ & f=R_1, .........(2)\\ & N-R_2-M g-m g / 2=0 .........(3) \end{aligned} $

The equations (1) and (3) give $R_2=0$ (as expected!) and $N=(M+m / 2) g$. In equilibrium, the net torque about any point is zero. Thus, the torque about the point $P$ for the left ladder is

$ M g(L / 2) \cos \theta+f L \sin \theta-N L \cos \theta=0 $ ..........(4)

Substitute $N=(M+m / 2) g$ in equation (4) and simplify to get

$ f=\left(\frac{M+m}{2}\right) g \cot \theta . $