A person of mass M is sitting on a swing to length L and swinging with an angular amplitude theta_0.

Q: Why is angular momentum conserved when the person stands up?

When the person stands up, the force they exert is internal to the system. The only external forces are gravity and the tension from the pivot. At the lowest point, gravity acts vertically downwards, passing directly through the pivot point. Therefore, the net external torque about the pivot is zero, allowing us to conserve angular momentum.

Q: Why does the velocity increase when the person stands up?

According to the conservation of angular momentum ($Mv_0L = Mv_1(L-l)$), if the radius of rotation decreases (from $L$ to $L-l$), the linear velocity must increase to keep the product constant. It's similar to an ice skater pulling their arms in to spin faster!

Q: Why is the work done by gravity negative?

As the person stands up, their center of mass moves upwards by a distance $l$. Since gravity acts downwards and the displacement is upwards, the angle between the force and displacement is $180^\circ$, making the work done by gravity negative ($-Mgl$).

Q: How did we get the initial velocity $v_0$?

The initial velocity $v_0$ is the maximum velocity of a simple pendulum at its lowest point. Using energy conservation from the extreme position (angle $\theta_0$) to the mean position, we get $v_0 = \sqrt{2gL(1-\cos\theta_0)}$. For small angles, $1-\cos\theta_0 \approx \theta_0^2/2$, which gives $v_0 = \theta_0\sqrt{gL}$.

Q: Why do we need the binomial approximation?

The expression for the new kinetic energy contains the term $(1 - l/L)^{-2}$. Since $l$ is much smaller than $L$, calculating this exactly is cumbersome. The binomial approximation $(1 - x)^{-n} \approx 1 + nx$ simplifies this complex fraction into a linear term, making the final algebraic manipulation elegant and straightforward.

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 $l (l << L)$, 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 problem of a person standing up on a swing is a classic, beautiful illustration of how multiple physical principles intertwine. It might seem like a simple playground scenario, but beneath the surface, it's a symphony of angular momentum, work, and energy!

### The Physics of the Playground

Imagine you are sitting on a swing of length $L$, swinging back and forth with an angular amplitude $	heta_0$. As you reach the lowest point of your trajectory, you suddenly stand up. Your center of mass shifts upwards by a small distance $l$. The question asks: how much work did you do to perform this action?

To solve this, we need to break the event into two distinct phases: the rapid standing motion at the lowest point, and the energy changes associated with it.

### The Pivot and the Conservation Law

When you stand up at the exact lowest point, the force you exert to lift yourself is an internal force. What about external forces? Gravity acts downwards, and the tension from the swing acts upwards. Both of these forces pass directly through the pivot point of the swing.

Because the line of action of these forces passes through the pivot, they exert **zero torque** about the pivot. With no net external torque, the **angular momentum of the system must be conserved**.

Let $v_0$ be your velocity just before standing, and $v_1$ be your velocity just after standing. 
Initial angular momentum: $L_i = M v_0 L$
Final angular momentum: $L_f = M v_1 (L - l)$

Equating them:
$$M v_0 L = M v_1 (L - l)$$
$$v_1 = v_0 \left( \frac{L}{L - l} ight)$$

Notice that since the radius decreased from $L$ to $L-l$, your velocity $v_1$ must be greater than $v_0$. You speed up just by standing!

### The Work-Energy Theorem

Now, let's find the work done. The Work-Energy Theorem states that the net work done on a system equals its change in kinetic energy:
$$W_{	ext{net}} = \Delta KE$$

The net work is the sum of the work done by gravity ($W_g$) and the work done by the person ($W_p$).
$$W_g + W_p = \frac{1}{2}Mv_1^2 - \frac{1}{2}Mv_0^2$$

As you stand up, your center of mass moves upwards by a distance $l$ against gravity. Therefore, the work done by gravity is negative:
$$W_g = -Mgl$$

Substituting this into our equation and isolating $W_p$:
$$W_p = Mgl + \frac{1}{2}M(v_1^2 - v_0^2)$$

### The Art of Approximation

Let's substitute our expression for $v_1$ into the work equation:
$$W_p = Mgl + \frac{1}{2}M \left[ \left( \frac{L}{L - l} ight)^2 v_0^2 - v_0^2 ight]$$
$$W_p = Mgl + \frac{1}{2}Mv_0^2 \left[ \left( 1 - \frac{l}{L} ight)^{-2} - 1 ight]$$

Here is where the magic happens. The problem states that $l \ll L$. This is a perfect invitation to use the **binomial approximation**: $(1 - x)^{-n} \approx 1 + nx$ for very small $x$.
$$\left( 1 - \frac{l}{L} ight)^{-2} \approx 1 + 2\left(\frac{l}{L}ight)$$

Substituting this back into our work equation:
$$W_p = Mgl + \frac{1}{2}Mv_0^2 \left[ \left( 1 + \frac{2l}{L} ight) - 1 ight]$$
$$W_p = Mgl + Mv_0^2 \frac{l}{L}$$

### The Final Synthesis

We are almost done, but we need to express $v_0$ in terms of the given variables. For a simple pendulum, the maximum velocity at the lowest point is related to the angular amplitude $	heta_0$:
$$v_0 = \omega_{	ext{max}} L = \left( 	heta_0 \sqrt{\frac{g}{L}} ight) L = 	heta_0 \sqrt{gL}$$

Squaring this gives $v_0^2 = 	heta_0^2 gL$. Let's plug this final piece into our work equation:
$$W_p = Mgl + M(	heta_0^2 gL) \frac{l}{L}$$
$$W_p = Mgl + Mgl 	heta_0^2$$
$$W_p = Mgl(1 + 	heta_0^2)$$

And there we have it! The work done by the person is $Mgl(1 + 	heta_0^2)$. This elegant result shows that the work done is not just the potential energy $Mgl$ required to lift the body, but also an additional term $Mgl	heta_0^2$ which provides the extra kinetic energy required to conserve angular momentum.

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Visualized Solution

$Visualizing the Swing$

$Conservation of Angular Momentum$

$Applying Angular Momentum Conservation$

$Finding New Velocity$

$Work-Energy Theorem$

$Work Done by Gravity$

$Isolating Work Done by Person$

$Substituting v_{1}$

$Binomial Approximation$

$Simplifying the Expression$

$Velocity of a Pendulum$

$Final Substitution$

$Conclusion$

The Sigma Insight: Conservation of Angular Momentum

The Physics of the Playground

The Pivot and the Conservation Law

The Work-Energy Theorem

The Art of Approximation

The Final Synthesis

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Swing

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Finding New Velocity

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Work-Energy Theorem

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Work Done by Gravity

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Isolating Work Done by Person

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting v1​

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Simplifying the Expression

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Velocity of a Pendulum

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

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