JEE Challenger
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PhyHardAdv 2024 • Paper 1

Impulse Imparted to Uniform Rod Connected to String

A thin uniform rod of length LL and certain mass is kept on a frictionless horizontal table with a massless string of length LL 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 PP is imparted to the rod at a distance x=L/nx = 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 nn is ______.

Question Diagram 1
Official Numerical Answer18

Topics & Concepts

System of Particles and Rotational MotionAngular MomentumMoment of Inertia

Step-by-Step Solution

To find the value of nn, we analyze the motion of the system (rod + massless string) immediately after the horizontal impulse PP is applied.

1. System Geometry and Parameters

  • Mass of the rod = mm
  • Length of the rod = LL
  • Length of the string = LL
  • Distance from the pivot O\text{O} to the inner end of the rod = LL
  • Distance from the pivot O\text{O} to the center of mass (CM\text{CM}) of the rod: rcm=L+L2=32Lr_{\text{cm}} = L + \frac{L}{2} = \frac{3}{2}L
  • Position of the point of application of impulse PP from the pivot O\text{O}: rP=rcm+x=32L+xr_P = r_{\text{cm}} + x = \frac{3}{2}L + x

2. Moment of Inertia about the Pivot O\text{O}

Using the parallel-axis theorem, the moment of inertia IOI_{\text{O}} of the rod about point O\text{O} is: IO=Icm+mrcm2=112mL2+m(32L)2I_{\text{O}} = I_{\text{cm}} + m r_{\text{cm}}^2 = \frac{1}{12}m L^2 + m\left(\frac{3}{2}L\right)^2 IO=112mL2+94mL2=(1+2712)mL2=73mL2I_{\text{O}} = \frac{1}{12}m L^2 + \frac{9}{4}m L^2 = \left(\frac{1 + 27}{12}\right) m L^2 = \frac{7}{3}m L^2

3. Impulse and Linear Momentum

Since the string is massless and flexible, it can only exert tension along its length (radially towards point O\text{O}). It cannot exert any transverse (perpendicular) force or impulse on the rod.

Therefore, the net transverse impulse on the rod is simply PP. By the impulse-momentum theorem for the center of mass in the transverse direction: P=mvcm  ⟹  vcm=PmP = m v_{\text{cm}} \implies v_{\text{cm}} = \frac{P}{m}

Given that the rod and string revolve together around point O\text{O} as a single rigid body with angular velocity ω\omega, the velocity of the center of mass is related to ω\omega by: vcm=ωrcm=ω(32L)v_{\text{cm}} = \omega r_{\text{cm}} = \omega \left(\frac{3}{2}L\right)

Equating the two expressions for vcmv_{\text{cm}}: ω(32L)=Pm  ⟹  ω=2P3mL\omega \left(\frac{3}{2}L\right) = \frac{P}{m} \implies \omega = \frac{2P}{3mL}

4. Angular Momentum Conservation about O\text{O}

The angular impulse imparted about the pivot O\text{O} is: ΔLO=P⋅rP=P(32L+x)\Delta L_{\text{O}} = P \cdot r_P = P \left(\frac{3}{2}L + x\right)

The angular momentum of the system after the impulse is: LO=IOω=(73mL2)(2P3mL)=149PLL_{\text{O}} = I_{\text{O}} \omega = \left(\frac{7}{3}m L^2\right) \left(\frac{2P}{3mL}\right) = \frac{14}{9}PL

Equating the angular impulse to the angular momentum: P(32L+x)=149PLP \left(\frac{3}{2}L + x\right) = \frac{14}{9}PL

Dividing both sides by PP: 32L+x=149L\frac{3}{2}L + x = \frac{14}{9}L

x=(149−32)L=(28−2718)L=L18x = \left(\frac{14}{9} - \frac{3}{2}\right) L = \left(\frac{28 - 27}{18}\right) L = \frac{L}{18}

Comparing with x=Lnx = \frac{L}{n}, we get: n=18n = 18