JEE Advanced 1997

LEVELJEE Main

Animated Solution for Physics - Rotational Motion: A rod of weight $W$ is supported by two parallel knife edges $A$ and $B$ and is in equilibrium in a horizontal position. The knives are at a distance $d$ from each other. The centre of mass of the rod is at distance $x$ from $A$ . The normal reaction on $A$ is ......... and on $B$ is ......... .

Q: What if the centre of mass was exactly in the middle?

If the CM is in the middle, then $x = d/2$. Substituting this into our formulas gives $N_A = \frac{d - d/2}{d}W = W/2$ and $N_B = \frac{d/2}{d}W = W/2$. The weight is distributed equally, which matches our intuition!

Q: What happens if the centre of mass is outside the knife edges?

If the CM is outside the supports (e.g., $x > d$ or $x < 0$), one of the normal reactions would calculate to be negative. A negative normal reaction means the support would have to pull *down* to keep the rod in equilibrium. Since knife edges can only push up, the rod would tip over!

Visualized Solution

$F r ee B o d y D ia g r am$

Let's draw the free body diagram of the rod.
The forces acting on the rod are its weight $W$ downwards at the centre of mass, and normal reactions $N_{A}$ and $N_{B}$ upwards at the knife edges $A$ and $B$ .

$D i s t an ces$

The distance between the knife edges is $d$ .
The centre of mass is at a distance $x$ from $A$ .
Therefore, its distance from $B$ is $d - x$ .

$R o t a t i o na l E q u i l ib r i u m$

For the rod to be in rotational equilibrium, the net torque about any point must be zero.
Let's calculate the torque about point $B$ .
$\sum τ_{B} = 0$

$T or q u e E q u a t i o nab o u tB$

Torque due to $N_{A}$ is clockwise (negative) and torque due to $W$ is counter-clockwise (positive).
$N_{A} \cdot d - W \cdot (d - x) = 0$

$S o l v in g f or N_{A}$

Rearranging the torque equation to solve for $N_{A}$ :
$N_{A} \cdot d = W (d - x)$
$N_{A} = \frac{d - x}{d} W$

$T r an s l a t i o na l E q u i l ib r i u m$

For the rod to be in translational equilibrium, the net vertical force must be zero.
$\sum F_{y} = 0$
$N_{A} + N_{B} = W$

$S o l v in g f or N_{B}$

Substitute the value of $N_{A}$ into the force equation to find $N_{B}$ :
$N_{B} = W - N_{A}$
$N_{B} = W - (\frac{d - x}{d}) W$

$F ina l R es u l t$

Simplifying the expression for $N_{B}$ :
$N_{B} = W (1 - \frac{d - x}{d}) = W (\frac{d - ( d - x )}{d}) = \frac{x}{d} W$
The normal reactions are $N_{A} = \frac{d - x}{d} W$ and $N_{B} = \frac{x}{d} W$ .

$T h e W a y F or w a r d$

What if the rod was not uniform and we didn't know the position of the centre of mass, but we knew the normal reactions?
We could use the same equations in reverse to find the centre of mass! This principle is used in weighing scales.

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The Sigma Insight: Dynamics of Rigid Body Rotation

Solution Diagram

Analyzing the Setup

Imagine a rod resting horizontally on two knife edges, much like a bridge supported by two pillars. The rod has a weight

W

, which acts entirely through its centre of mass (CM). The knife edges, located at points

A

and

B

, push back against the rod to keep it from falling. These upward pushes are the normal reactions,

N_{A}

and

N_{B}

We are given that the distance between the two knife edges is

d

. The centre of mass is located at a distance

x

from knife edge

A

. By simple geometry, the distance from the centre of mass to knife edge

B

must be

d - x

The Master Equation

Rotational Equilibrium

For the rod to remain perfectly horizontal and not tilt, it must be in rotational equilibrium. This means the net torque (turning effect) about any point on the rod must be exactly zero.

To make our calculations elegant, let's choose point

B

as our pivot for calculating torque. Why

B

? Because the normal reaction

N_{B}

passes directly through this point, its perpendicular distance is zero, and thus its torque is zero. This clever choice eliminates one unknown from our equation!

The forces creating torque about point

B

are: 1. The normal reaction

N_{A}

pushing upwards at a distance

d

. This creates a clockwise torque. 2. The weight

W

pulling downwards at a distance

d - x

. This creates a counter-clockwise torque.

Equating the magnitudes of these opposing torques:

N_{A} \cdot d = W \cdot (d - x)

Solving for the First Reaction

From our torque equation, isolating

N_{A}

is straightforward:

N_{A} = \frac{d - x}{d} W

This result makes intuitive sense. If the centre of mass is closer to

B

(meaning

x

is large and

d - x

is small), the reaction at

A

will be smaller.

Translational Equilibrium

Now, let's look at the vertical forces. For the rod to not accelerate upwards or downwards, it must be in translational equilibrium. The total upward force must perfectly balance the total downward force.

The upward forces are

N_{A}

and

N_{B}

, and the downward force is

W

N_{A} + N_{B} = W

Final Calculation

We already know

N_{A}

, so we can easily find

N_{B}

by substituting it into our force balance equation:

N_{B} = W - N_{A}

N_{B} = W - (\frac{d - x}{d}) W

Let's factor out

W

and simplify the fraction:

N_{B} = W (1 - \frac{d - x}{d})

N_{B} = W (\frac{d - ( d - x )}{d})

N_{B} = \frac{x}{d} W

And there we have it! The normal reactions are $N_{A} = \frac{d - x}{d} W$ and $N_{B} = \frac{x}{d} W$ . Notice how the reaction at each support is proportional to the distance of the centre of mass from the other support.

Common Traps & FAQs

Why did we take torque about point B? Could we take it about point A or the centre of mass?

You can take torque about ANY point! The rod is in equilibrium, so net torque is zero everywhere. We chose point B because the torque of

N_{B}

becomes zero, giving us an equation with only one unknown (

N_{A}

). Taking torque about A would give

N_{B}

directly. Taking it about the CM would give an equation with both

N_{A}

and

N_{B}

Does the thickness or length of the rod outside the knife edges matter?

No, it doesn't matter for this specific problem. The entire weight of the rod can be considered to act at its centre of mass. As long as we know the position of the CM relative to the supports, the overhanging parts don't change the equations.

What if the centre of mass was exactly in the middle?

If the CM is in the middle, then

x = d /2

. Substituting this into our formulas gives

N_{A} = \frac{d - d /2}{d} W = W /2

and

N_{B} = \frac{d /2}{d} W = W /2

. The weight is distributed equally, which matches our intuition!

Why are the normal reactions considered upwards?

The knife edges are supporting the rod, preventing it from falling due to gravity. Therefore, they must exert an upward push on the rod.

What happens if the centre of mass is outside the knife edges?

If the CM is outside the supports (e.g.,

x > d

x < 0

), one of the normal reactions would calculate to be negative. A negative normal reaction means the support would have to pull *down* to keep the rod in equilibrium. Since knife edges can only push up, the rod would tip over!

Visualized Solution

$F r ee B o d y D ia g r am$

$D i s t an ces$

$R o t a t i o na l E q u i l ib r i u m$

$T or q u e E q u a t i o nab o u tB$

$S o l v in g f or N_{A}$

$T r an s l a t i o na l E q u i l ib r i u m$

$S o l v in g f or N_{B}$

$F ina l R es u l t$

$T h e W a y F or w a r d$

The Sigma Insight: Dynamics of Rigid Body Rotation

Analyzing the Setup

The Master Equation

Solving for the First Reaction

Translational Equilibrium

Final Calculation

Similar Questions

A rigid massless rod of length $3 l$ has two masses attached at each end as shown in the figure. The rod is pivoted at point $P$ on the horizontal axis (see figure). When released from initial horizontal position, its instantaneous angular acceleration will be

A uniform rod of length $l$ is being rotated in a horizontal plane with a constant angular speed about an axis passing through one of its ends. If the tension generated in the rod due to rotation is $T (x)$ at a distance $x$ from the axis, then which of the following graphs depicts it most closely?

Visualized Solution

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The Sigma Insight: Dynamics of Rigid Body Rotation

Analyzing the Setup

The Master Equation

Solving for the First Reaction

Translational Equilibrium

Final Calculation

Similar Questions

$F r ee B o d y D ia g r am$

$D i s t an ces$

$R o t a t i o na l E q u i l ib r i u m$

$T or q u e E q u a t i o nab o u tB$

$S o l v in g f or N_{A}$

$T r an s l a t i o na l E q u i l ib r i u m$

$S o l v in g f or N_{B}$

$F ina l R es u l t$

$T h e W a y F or w a r d$