Three particles A, B and C, each of mass m, are connected to each other by three massless rigid rods

Q: Why do we only consider the centripetal acceleration for part (a)?

In part (a), the body is rotating with a *constant* angular velocity $\omega$. Because $\omega$ is constant, there is no angular acceleration ($\alpha = 0$), and thus no tangential acceleration. The only acceleration the Center of Mass has is the centripetal acceleration required to keep it moving in a circle.

Q: How did we find the perpendicular distance for the torque of force $F$?

The force $F$ is applied along the side $BC$. The perpendicular distance from the hinge $A$ to the line $BC$ is simply the height (or altitude) of the equilateral triangle. Using basic geometry, the height of an equilateral triangle of side $l$ is $\frac{\sqrt{3}}{2}l$.

Q: Why is the moment of inertia $2ml^2$ and not something else?

The moment of inertia is calculated about the hinge $A$. The system consists of three point masses. The mass at $A$ is at a distance of $0$ from the axis, so it contributes nothing. The masses at $B$ and $C$ are each at a distance $l$ from $A$. Therefore, $I_A = m(0)^2 + m(l)^2 + m(l)^2 = 2ml^2$.

Q: Why does the tangential acceleration of the CM point purely in the x-direction?

The Center of Mass lies exactly on the y-axis (the median passing through $A$). Tangential acceleration is always perpendicular to the position vector from the axis of rotation. Since the position vector is along the y-axis, the perpendicular direction is strictly along the x-axis.

Q: Why is the x-component of the hinge force negative?

When the force $F$ is applied to the right (positive x-direction), it causes the CM to accelerate to the right. However, the acceleration $a_x = \frac{F}{4m}$ requires a net force of $\frac{3F}{4}$. Since the applied force $F$ is larger than the required net force, the hinge must pull back to the left (negative x-direction) with a force of $\frac{F}{4}$ to keep the physics balanced!

Question

Three particles $A$, $B$ and $C$, each of mass $m$, are connected toeach other by three massless rigid rods to form a rigid , equilateral triangular body of side $l$. This body is placed on a horizontal frictionless table ($x$-$y$ plane) and is hinged to it at the point $A$, so that it can move without friction about the vertical axis through $A$ (see figure). The body is set into rotational motion on the table about $A$ with a constant angular velocity $\omega$ .
(a) Find the magnitude of the horizontal force exerted by the hinge on the body.
(b) At time $T$ , when the side $BC$ is parallel to the $X$-axis, a force $F$ is applied on $B$ along $BC$ (as shown). Obtain the $x$-component and the $y$-component of the force exerted by the hinge on the body, immediately after time $T$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are looking down at a perfectly smooth, frictionless table. On this table lies a rigid equilateral triangle formed by three massless rods, with a point mass $m$ at each of its three vertices: $A$, $B$, and $C$. The entire structure is pinned to the table at vertex $A$, allowing it to spin freely like a fidget spinner.

Before we dive into the forces, we must locate the heart of this system: the **Center of Mass (CM)**. Because the triangle is equilateral and the masses are identical, the CM lies exactly on the median passing through the hinge $A$. Using basic geometry, the distance $r$ from the hinge $A$ to the CM is given by:

$r = \frac{l}{\sqrt{3}}$

This geometric insight is the foundation for everything that follows.

### The Centripetal Reality

In part (a) of the problem, the triangular body is already spinning with a constant angular velocity $\omega$. Think about what this means for the Center of Mass. It is traveling in a perfect circle of radius $r$ around the hinge $A$.

Whenever an object moves in a circle, it demands a centripetal force to keep it from flying off in a straight line. The centripetal acceleration $a_c$ of the CM is directed straight towards the hinge $A$ and has a magnitude of:

$a_c = r\omega^2 = \frac{l}{\sqrt{3}}\omega^2$

Since the table is completely frictionless, what is providing this inward pull? It can only be the hinge itself! According to Newton's Second Law, the net force must equal the total mass of the system ($3m$) multiplied by its acceleration. Therefore, the force exerted by the hinge is:

$F_{	ext{hinge}} = (3m) a_c = (3m) \left(\frac{l}{\sqrt{3}}\omega^2ight) = \sqrt{3}ml\omega^2$

This force acts purely along the median (which we can call the y-axis), pulling the CM towards $A$.

### The Sudden Twist

Now, let's move to part (b). At a specific moment $T$, when the side $BC$ is perfectly horizontal (parallel to the x-axis), a new force $F$ is suddenly applied at vertex $B$, pointing directly towards $C$.

This force $F$ is trying to twist the triangle, creating a torque about the hinge $A$. To calculate this torque, we need the perpendicular distance from the line of action of the force to the hinge. This distance is simply the altitude of the equilateral triangle:

$d = \frac{\sqrt{3}}{2}l$

Thus, the torque $	au_A$ generated by the applied force is:

$	au_A = F \cdot \frac{\sqrt{3}}{2}l$

### The Resistance to Rotation

To find out how much this torque accelerates the rotation, we need the system's moment of inertia about the hinge $A$. The mass at $A$ is right on the axis of rotation, so its distance is zero. The masses at $B$ and $C$ are each at a distance $l$ from $A$.

$I_A = m(0)^2 + m(l)^2 + m(l)^2 = 2ml^2$

Using Newton's Second Law for rotation ($	au_A = I_A \alpha$), we can find the instantaneous angular acceleration $\alpha$:

$\alpha = \frac{	au_A}{I_A} = \frac{F \frac{\sqrt{3}}{2}l}{2ml^2} = \frac{\sqrt{3}F}{4ml}$

### The Tangential Shift

Because the entire body now has an angular acceleration $\alpha$, the Center of Mass will experience a tangential acceleration. Since the CM lies exactly on the y-axis, its tangential acceleration will be perfectly horizontal, pointing along the x-axis.

$a_x = r\alpha = \left(\frac{l}{\sqrt{3}}ight) \left(\frac{\sqrt{3}F}{4ml}ight) = \frac{F}{4m}$

This is a beautifully simple result. The complex geometry has canceled out, leaving us with a clean expression for the horizontal acceleration of the CM.

### Balancing the Forces

Now comes the grand finale. We must apply Newton's Second Law for linear motion in both the x and y directions to find the new components of the hinge force, $F_x$ and $F_y$.

Let's look at the x-direction. The total external force consists of the applied force $F$ and the x-component of the hinge force $F_x$. This net force must equal the total mass ($3m$) times the horizontal acceleration $a_x$:

$F_{	ext{net}, x} = F + F_x = (3m) a_x = (3m) \left(\frac{F}{4m}ight) = \frac{3F}{4}$

Solving for $F_x$, we get:

$F_x = \frac{3F}{4} - F = -\frac{F}{4}$

The negative sign is crucial! It tells us that the hinge is actually pushing back to the left, opposing the applied force $F$, to keep the physics perfectly balanced.

Finally, what about the y-direction? The newly applied force $F$ is purely horizontal, so it has absolutely no effect on the vertical forces. The Center of Mass still requires its centripetal force to maintain its circular path. Therefore, the y-component of the hinge force remains exactly the same as we calculated in part (a):

$F_y = \sqrt{3}ml\omega^2$

**And there we have it! The complete force exerted by the hinge has been successfully decoded.**

Visualized Solution

$P os i t i o n o f C e n t r e o f M a ss$

$C e n t r i p e t a l A cce l er a t i o n o f C M$

$H in g e F or ce B e f or e F i s A ppl i e d$

$A ppl i c a t i o n o f F or ce F$

$M o m e n t o f I n er t ia ab o u t A$

$A n g u l a r A cce l er a t i o n α$

$T an g e n t ia l A cce l er a t i o n o f C M$

$H in g e F or ce in x - d i r ec t i o n$

$H in g e F or ce in y - d i r ec t i o n$

The Sigma Insight: Dynamics of Rigid Body Rotation

Analyzing the Setup

The Centripetal Reality

The Sudden Twist

The Resistance to Rotation

The Tangential Shift

Balancing the Forces

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Position of Centre of Mass

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Centripetal Acceleration of CM

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Hinge Force Before F is Applied

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Application of Force F

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Moment of Inertia about A

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Angular Acceleration α

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Tangential Acceleration of CM

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Hinge Force in x−direction

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Hinge Force in y−direction