A person of mass M is sitting on a swing to length L and swinging with an angular amplitude θ_0. If

Q: Wait, why is angular momentum conserved, but not linear momentum?

Linear momentum is not conserved because there is a net external force acting on the system: the tension from the pivot and gravity. However, at the lowest point, both these forces pass through or are parallel to the position vector from the pivot, meaning they exert zero torque about the pivot. Zero external torque means angular momentum is conserved!

Q: Why does the person's speed increase when they stand up?

Think of an ice skater pulling their arms in. By standing up, the person reduces their distance from the pivot (their moment of inertia decreases). To conserve angular momentum ($L = mvr$), a decrease in radius $r$ must result in an increase in velocity $v$.

Q: Why is the work done by gravity negative?

Work is the dot product of force and displacement. Gravity acts downwards, but when the person stands up, their center of mass moves upwards by a distance $l$. Since the force and displacement are in opposite directions, the work done is negative ($-Mgl$).

Q: Why do we use the binomial approximation here?

The problem states that $l \ll L$ (the distance the person stands up is much smaller than the length of the swing). This allows us to use the approximation $(1 + x)^n \approx 1 + nx$ for very small $x$. It turns a complex algebraic fraction into a simple linear term, making the calculation elegant and solvable!

Q: Why can't we just use energy conservation for the whole process?

We *are* using the Work-Energy theorem, which is the generalized form of energy conservation! However, mechanical energy is NOT conserved during the act of standing up because the person's muscles are doing internal non-conservative work to lift their body against both gravity and the centrifugal force. That's exactly the work we are calculating!

Question

A person of mass $M$ is sitting on a swing to length $L$ and swinging with an angular amplitude $	heta_0$. If the person stands up when the swing passes through its lowest point, the work done by him, assuming that his centre of mass moves by a distance $l(l \ll L)$, is close to

Options:

(A) $Mgl (1 - 	heta_0^2)$

(B) $Mgl (1 + 	heta_0^2)$

(C) $Mgl$

(D) $Mgl \left(1 + \frac{	heta_0^2}{2}ight)$

Sigma Percentile · Accepted Answer

### The Physics of the Playground

Imagine you are back on the playground, sitting on a swing. You know intuitively that to go higher, you need to "pump" the swing by standing up and sitting down at specific points in the arc. But how much energy does that actually take?

In this classic JEE Advanced problem, we are going to calculate the exact mechanical work your muscles must do to stand up right as the swing passes through its lowest point.

### The Conservation of Angular Momentum

As you swing down to the lowest point, you have a maximum velocity $v_0$. At this exact instant, you suddenly stand up. What happens to your speed?

To answer this, we must look at the forces. Gravity acts downwards, and the tension from the swing's chains acts upwards. Both of these forces pass directly through the pivot point (or are parallel to the position vector). Because there is zero external torque acting about the pivot, **angular momentum is conserved**.

Let your initial distance from the pivot be $L$, and your final distance be $L - l$. Equating the initial and final angular momentum:

$$M v_0 L = M v_1 (L - l)$$

Rearranging this gives us your new, faster velocity $v_1$:

$$v_1 = v_0 \frac{L}{L - l}$$

By standing up, you decreased your moment of inertia, which forced your velocity to increase!

### The Work-Energy Theorem

Now, let's calculate the work. The Work-Energy Theorem states that the total work done by all forces equals the change in kinetic energy.

$$W_{	ext{gravity}} + W_{	ext{person}} = \Delta K$$

When you stand up, your center of mass moves upwards by a distance $l$. Gravity is fighting against you, so it does negative work: $W_{	ext{gravity}} = -Mgl$.

$$-Mgl + W_{	ext{person}} = \frac{1}{2}Mv_1^2 - \frac{1}{2}Mv_0^2$$

Isolating the work done by the person:

$$W_{	ext{person}} = Mgl + \frac{1}{2}M(v_1^2 - v_0^2)$$

### The Binomial Approximation

Let's substitute our expression for $v_1$ into the work equation:

$$W_{	ext{person}} = Mgl + \frac{1}{2}Mv_0^2 \left[ \left(\frac{L}{L - l}ight)^2 - 1 ight]$$

This looks messy, but the problem gives us a crucial hint: $l \ll L$. This is an invitation to use the binomial approximation, $(1 + x)^n \approx 1 + nx$.

First, rewrite the fraction:

$$\left(\frac{L}{L - l}ight)^2 = \left(1 - \frac{l}{L}ight)^{-2}$$

Applying the approximation:

$$\left(1 - \frac{l}{L}ight)^{-2} \approx 1 - 2\left(-\frac{l}{L}ight) = 1 + \frac{2l}{L}$$

Substitute this beautiful simplification back into our work equation:

$$W_{	ext{person}} = Mgl + \frac{1}{2}Mv_0^2 \left[ \left(1 + \frac{2l}{L}ight) - 1 ight]$$

$$W_{	ext{person}} = Mgl + Mv_0^2 \frac{l}{L}$$

### The Energy of the Swing

We are almost there, but we need to find $v_0^2$. We can find this by looking at the swing's journey from its highest point (amplitude $	heta_0$) to the lowest point.

By conservation of mechanical energy, the kinetic energy at the bottom equals the gravitational potential energy lost from the top:

$$\frac{1}{2}Mv_0^2 = MgL(1 - \cos	heta_0)$$

For a small angular amplitude, we can use the small-angle approximation $1 - \cos	heta_0 \approx \frac{	heta_0^2}{2}$:

$$\frac{1}{2}Mv_0^2 = MgL \left(\frac{	heta_0^2}{2}ight)$$

$$v_0^2 = gL	heta_0^2$$

### The Final Synthesis

Finally, substitute $v_0^2$ into our simplified work equation:

$$W_{	ext{person}} = Mgl + M(gL	heta_0^2) \frac{l}{L}$$

The $L$ cancels out perfectly:

$$W_{	ext{person}} = Mgl + Mgl	heta_0^2$$

Factoring out $Mgl$, we arrive at our elegant final answer:

$$W_{	ext{person}} = Mgl(1 + 	heta_0^2)$$

This result is profound. The work you do is not just $Mgl$ (the work required to lift your body at rest). You must do *extra* work ($Mgl	heta_0^2$) to fight against the centrifugal force pulling you outwards as you swing!

Select Answer:

Visualized Solution

Visualizing the Setup

The Act of Standing

Conservation of Angular Momentum

Angular Momentum Equation

Velocity After Standing

Work-Energy Theorem

Setting Up the Work Equation

Work Done by Gravity

Substituting Velocity

Binomial Approximation

Simplifying the Work Equation

Finding Initial Velocity

Energy Conservation

Small Angle Approximation

The Final Answer

The Sigma Insight: Conservation of Angular Momentum

The Physics of the Playground

The Conservation of Angular Momentum

The Work-Energy Theorem

The Binomial Approximation

The Energy of the Swing

The Final Synthesis

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