A thin uniform rod of length L and certain mass is kept on a frictionless horizontal table with a ma

Q: Why do we assume the rod and string rotate as a rigid body?

Because the problem states they 'revolve together... remaining aligned'. If they weren't aligned, the rod would have an additional angular velocity relative to the string.

Q: Wait, doesn't the pivot exert a reaction impulse?

Ah, this is a great question! The pivot *can* exert a reaction impulse along the string (radial direction) to provide centripetal force, but in the tangential direction (perpendicular to the rod), the pivot exerts zero impulse because the flexible string cannot transmit shear force. Thus, the tangential linear momentum is conserved with respect to the applied impulse $P$.

Q: How did you calculate the angular momentum $L_O$?

We used the parallel axis theorem concept for angular momentum: $L_O = I_{cm} \omega + \vec{r}_{cm} \times \vec{p}_{cm}$. Since the velocity is purely tangential, this simplifies to $I_{cm} \omega + m v_{cm} r_{cm}$.

Q: Why is the distance of CM from O taken as $3L/2$?

The string has length $L$, and the rod has length $L$. The center of mass of the uniform rod is exactly in its middle, which is $L/2$ from its connected end. So, the total distance from the pivot O is $L + L/2 = 3L/2$.

Question

A thin uniform rod of length L and certain mass is kept on a frictionless horizontal table with a massless string of length L fixed to one end (top view is shown in the figure). The other end of the string is pivoted to a point O. If a horizontal impulse P is imparted to the rod at a distance x = L/n from the mid-point of the rod (see figure), then the rod and string revolve together around the point O, with the rod remaining aligned with the string. In such a case, the value of n is _____.

Sigma Percentile · Accepted Answer

The problem of a rod and string revolving together after an impulse is a beautiful exploration of rigid body dynamics and conservation laws. It tests our ability to seamlessly connect the linear and angular domains. Let's dive into the physics of this elegant setup!

### Analyzing the Setup

Imagine you are looking down at a frictionless table. A uniform rod of length $L$ is attached to a massless string, also of length $L$, which is anchored at a pivot point $O$.

Suddenly, a sharp horizontal impulse $P$ is applied to the rod at a specific distance $x$ from its center of mass. The problem states a crucial constraint: **the rod and string revolve together, remaining perfectly aligned**.

This constraint is our golden key. It tells us that the entire system behaves as a single rigid body rotating about the pivot $O$. There is no bending at the joint between the string and the rod. Because of this, every point on the rod shares the same angular velocity, $\omega$.

Let's locate the center of mass (CM) of the rod. Since the string has length $L$ and the rod has length $L$, the CM of the uniform rod is exactly in its middle. 
The distance of the CM from the pivot $O$ is:

$$ r_{cm} = L + \frac{L}{2} = \frac{3L}{2} $$

Because the rod rotates as a rigid body, the linear velocity of its center of mass is directly tied to the angular velocity:

$$ v_{cm} = r_{cm} \omega = \frac{3L}{2} \omega $$

### The Master Equations

When the impulse $P$ strikes the rod, it imparts both linear and angular momentum to the system. We need to analyze both domains.

**1. Linear Momentum**
According to the impulse-momentum theorem, the net linear impulse applied to a system equals its change in linear momentum. Since the string can only exert a radial force (tension) and cannot transmit any tangential shear force, the pivot exerts zero tangential impulse. Thus, the applied impulse $P$ is entirely responsible for the tangential velocity of the rod.

$$ P = m v_{cm} $$

Substituting our expression for $v_{cm}$, we get:

$$ P = m \left( \frac{3L}{2} \omega ight) $$

From this, we can isolate the angular velocity $\omega$:

$$ \omega = \frac{2P}{3mL} $$

**2. Angular Momentum**
Now, let's look at the rotational effect. The angular impulse about the pivot $O$ must equal the change in the system's angular momentum about $O$.

The impulse $P$ is applied at a distance $x$ from the CM. Therefore, its total perpendicular distance from the pivot $O$ is $r_{cm} + x = \frac{3L}{2} + x$.
The angular impulse is simply the force (or impulse) multiplied by this lever arm:

$$ 	ext{Angular Impulse} = P \left( \frac{3L}{2} + x ight) $$

Next, we calculate the final angular momentum of the rod about $O$. Using the parallel axis theorem concept for angular momentum, it is the sum of the spin angular momentum about the CM and the orbital angular momentum of the CM about $O$:

$$ L_O = I_{cm} \omega + m v_{cm} r_{cm} $$

We know the moment of inertia of a uniform rod about its center is $I_{cm} = \frac{mL^2}{12}$. Let's plug in all our knowns:

$$ L_O = \frac{mL^2}{12} \omega + m \left( \frac{3L}{2} \omega ight) \left( \frac{3L}{2} ight) $$

$$ L_O = \left( \frac{1}{12} + \frac{9}{4} ight) mL^2 \omega $$

Finding a common denominator (12), we get:

$$ L_O = \left( \frac{1 + 27}{12} ight) mL^2 \omega = \frac{28}{12} mL^2 \omega = \frac{7}{3} mL^2 \omega $$

### Final Calculation

We have our angular impulse and our angular momentum. By the principle of angular impulse and momentum, we equate them:

$$ P \left( \frac{3L}{2} + x ight) = \frac{7}{3} mL^2 \omega $$

This is where the magic happens. We substitute the expression for $\omega$ that we found from the linear momentum equation:

$$ P \left( \frac{3L}{2} + x ight) = \frac{7}{3} mL^2 \left( \frac{2P}{3mL} ight) $$

Notice how beautifully the mass $m$ and one power of $L$ cancel out on the right side. Even the impulse $P$ cancels out from both sides!

$$ \frac{3L}{2} + x = \frac{14}{9} L $$

Now, it's just a matter of simple algebra to solve for $x$:

$$ x = \frac{14}{9} L - \frac{3}{2} L $$

$$ x = \left( \frac{28 - 27}{18} ight) L = \frac{L}{18} $$

The problem states that $x = \frac{L}{n}$. Comparing our result, we can confidently conclude that:

$$ n = 18 $$

This problem is a fantastic reminder of why we must trust the fundamental conservation laws. By methodically applying linear and angular impulse equations, even the most complex-looking rigid body dynamics problems unravel into elegant algebraic solutions.

Enter Numerical Value:

Visualized Solution

$P, L, x$

$v_{c m} = r_{c m} ω$

$r_{c m} = L + \frac{L}{2} = \frac{3 L}{2}$

$P = Δ p = m v_{c m}$

$ω = \frac{2 P}{3 m L}$

$\int τ d t = Δ L_{O}$

$Angular Impulse = P (\frac{3 L}{2} + x)$

$L_{O} = I_{c m} ω + m v_{c m} r_{c m}$

$L_{O} = \frac{7}{3} m L^{2} ω$

$P (\frac{3 L}{2} + x) = \frac{7}{3} m L^{2} ω$

$P (\frac{3 L}{2} + x) = \frac{14}{9} P L$

$x = \frac{L}{18}$

$What if...$

The Sigma Insight: Torque and Angular Momentum

Analyzing the Setup

The Master Equations

Final Calculation

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Enter Numerical Value:

Visualized Solution

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } vcm​=rcm​ω

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } rcm​=L+2L​=23L​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P=Δp=mvcm​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ω=3mL2P​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∫τdt=ΔLO​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Angular Impulse=P(23L​+x)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } LO​=Icm​ω+mvcm​rcm​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } LO​=37​mL2ω

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(23L​+x)=37​mL2ω

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(23L​+x)=914​PL

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=18L​

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