Rotational Equilibrium Questions in English

(A) Let $L$ be the length of the iron bar. The weight $W$ of the bar acts at its center of mass,which is at a distance of $L/2$ from the end on the ground.
Let $R$ be the reaction force exerted by the person's shoulder on the bar. The bar is in rotational equilibrium.
Taking the torque about the point of contact on the ground:
The torque due to the weight $W$ is $\tau_W = W \cdot (L/2) \cos \theta$ (acting in a clockwise direction).
The torque due to the reaction force $R$ is $\tau_R = R \cdot L \cos \theta$ (acting in a counter-clockwise direction).
For rotational equilibrium,the net torque about the point of contact on the ground must be zero:
$\sum \tau = 0$
$R \cdot L \cos \theta = W \cdot (L/2) \cos \theta$
Dividing both sides by $L \cos \theta$ (assuming $\cos \theta \neq 0$):
$R = \frac{W}{2}$
Thus,the weight experienced by the person is $\frac{W}{2}$.

(B) For the rod to be in rotational equilibrium,the net torque about the pivot point (at $40 \ cm$ mark) must be zero.
Let $m$ be the mass to be suspended at the $90 \ cm$ mark.
The weight of the rod acts at its center of mass,which is at the $50 \ cm$ mark.
The distance of the $400 \ g$ mass from the pivot is $40 \ cm - 10 \ cm = 30 \ cm$.
The distance of the rod's center of mass from the pivot is $50 \ cm - 40 \ cm = 10 \ cm$.
The distance of the unknown mass $m$ from the pivot is $90 \ cm - 40 \ cm = 50 \ cm$.
Taking the pivot as the reference point,the torque balance equation is:
$(400 \ g \times 30 \ cm) = (250 \ g \times 10 \ cm) + (m \times 50 \ cm)$
$12000 = 2500 + 50m$
$50m = 12000 - 2500$
$50m = 9500$
$m = \frac{9500}{50} = 190 \ g$

(B) The system is in translational and rotational equilibrium.
For translational equilibrium,the sum of upward forces equals the downward force (weight of the rod):
$T_1 + T_2 = mg = 10 \ kg \times 10 \ m/s^2 = 100 \ N$ (Equation $1$)
For rotational equilibrium,the net torque about any point must be zero. Let us take the torque about the point where $T_2$ is applied:
$\tau_{\text{net}} = 0$
$T_1 \times \ell - mg \times \frac{\ell}{2} = 0$
$T_1 \times \ell = 100 \times \frac{\ell}{2}$
$T_1 = 50 \ N$

(C) For a uniform meter scale,the weight of the scale acts at its center of gravity,which is at the $50 \ cm$ mark.
Let the weight of the scale be $W = mg$.
The pivot point is at the $30 \ cm$ mark.
According to the principle of moments,the sum of clockwise moments about the pivot must equal the sum of counter-clockwise moments about the pivot.
Counter-clockwise moment: $60 \ N \times (30 \ cm - 10 \ cm) = 60 \times 20 = 1200 \ N \cdot cm$.
Clockwise moments: $W \times (50 \ cm - 30 \ cm) + 10 \ N \times (80 \ cm - 30 \ cm) = W \times 20 + 10 \times 50 = 20W + 500$.
Equating the moments: $1200 = 20W + 500$.
$20W = 1200 - 500 = 700$.
$W = 35 \ N$.
Since $W = mg$ and $g = 10 \ m/s^2$,we have $m \times 10 = 35$.
$m = 3.5 \ kg$.

(B) Let the mass of the meter stick be $m$. The center of mass of the uniform meter stick is at the $50 \ cm$ mark.
When the stick is balanced at the $45 \ cm$ mark,the pivot is at $45 \ cm$.
The distance of the coin from the pivot is $|45 \ cm - 12 \ cm| = 33 \ cm$.
The distance of the center of mass of the stick from the pivot is $|50 \ cm - 45 \ cm| = 5 \ cm$.
For rotational equilibrium,the net torque about the pivot must be zero:
$\tau_{\text{net}} = 0$
$(10 \ g) \times (33 \ cm) = (m \ g) \times (5 \ cm)$
$330 = 5m$
$m = \frac{330}{5} = 66 \ g$
Therefore,the mass of the meter stick is $66 \ g$.

(D) For the rod to be in rotational equilibrium,the sum of torques about any point must be zero.
Taking the torque about the knife-edge $B$:
$\sum \tau_B = 0$
$N_A \cdot r - W \cdot (r - x) = 0$
Where $N_A$ is the normal reaction at $A$.
$N_A \cdot r = W(r - x)$
$N_A = \frac{W(r - x)}{r}$

(A) The centre of gravity of a metre scale is at the $50 \text{ cm}$ mark. The wedge is placed at this point.
The distance of the weight '$W$' from the wedge is $50 \text{ cm} - 20 \text{ cm} = 30 \text{ cm}$.
The distance of the $25 \text{ g-wt}$ weight from the wedge is $74 \text{ cm} - 50 \text{ cm} = 24 \text{ cm}$.
For the scale to remain horizontal, the clockwise moment must equal the anticlockwise moment about the pivot (wedge):
$W \times 30 \text{ cm} = 25 \text{ g-wt} \times 24 \text{ cm}$
$W = \frac{25 \times 24}{30} \text{ g-wt}$
$W = \frac{600}{30} \text{ g-wt} = 20 \text{ g-wt}$

(C) Let $M$ be the mass of the metre stick. The centre of mass of the metre stick is at the $50.0 \,cm$ mark.
When the stick is balanced at the $46.0 \,cm$ mark, the torque due to the weight of the stick must be balanced by the torque due to the coins.
The weight of the four coins is $W_c = 4 \times 2 \,g = 8 \,g$. This acts at the $10.0 \,cm$ mark.
The distance of the coins from the pivot $(46.0 \,cm)$ is $d_1 = 46.0 \,cm - 10.0 \,cm = 36.0 \,cm$.
The weight of the metre stick $W_s = M \,g$ acts at its centre of mass, which is at the $50.0 \,cm$ mark.
The distance of the centre of mass from the pivot is $d_2 = 50.0 \,cm - 46.0 \,cm = 4.0 \,cm$.
For rotational equilibrium, the clockwise torque must equal the counter-clockwise torque:
$M \,g \times d_2 = W_c \times d_1$
$M \,g \times 4.0 \,cm = 8 \,g \times 36.0 \,cm$
$4 \,M = 8 \times 36$
$M = 2 \times 36 = 72 \,g$.

(C) The forces acting on the ball are the tension $T$ in the string and the gravitational force $mg$ acting downwards.
Let the centre of the horizontal circle be $O$. The position vector $\vec{r}$ of the ball relative to $O$ lies in the horizontal plane.
The gravitational force $\vec{F}_g = m\vec{g}$ acts vertically downwards through the ball.
The torque $\vec{\tau}$ about point $O$ is given by $\vec{\tau} = \vec{r} \times \vec{F}$.
Since the force of gravity $\vec{F}_g$ is parallel to the vertical axis passing through $O$,and the position vector $\vec{r}$ is horizontal,the torque due to gravity is $\vec{\tau}_g = \vec{r} \times (m\vec{g})$. The magnitude is $mgr \sin(90^\circ) = mgr$,but the direction is tangential to the circle.
However,the tension force $\vec{T}$ acts along the string towards the point of suspension. The line of action of the tension force passes through the point of suspension,but not through $O$.
Crucially,for a particle moving in a horizontal circle with constant angular velocity,the net torque about the centre $O$ must be zero because the angular momentum $\vec{L} = \vec{r} \times \vec{p}$ is constant in magnitude and direction (the vector $\vec{L}$ points along the vertical axis). Since $\frac{d\vec{L}}{dt} = \vec{\tau}_{net} = 0$,the net torque about $O$ is zero.