Question

69. A thin rigid insulating ring of mass m = 0.1 kg and radius R = 1 m is free to rotate about a fixed vertical axis O, perpendicular to the plane of the ring (see Figure). Ring uniformly charged in length and its charge is Q = 100$\pi$ C. A very small piece of the ring in the area of point A cut so that you will get a gap of length $l$ =0.1 m. A uniform electric field E = 4 N/C is applied parallel to 'l' and the ring is released from rest.

The maximum angular velocity $\omega$ of the ring in the subsequent motion will be :

Answer 1

20\sqrt{2} rad/s

Answer 2

Solution:

1. Find the missing charge and effective dipole moment:

   • Charge per unit length on the ring: $$\lambda=\frac{Q}{2\pi R}$$
   • Charge missing due to the gap of length $l$: $$\Delta q=\lambda\,l=\frac{Ql}{2\pi R}$$
   • This missing charge at a distance $R$ creates an effective dipole moment: $$p=\Delta q\cdot R=\frac{Ql}{2\pi}\,.$$

2. Energy considerations:
   When the ring rotates, the change in electric potential energy of the dipole in the field $E$ is converted into rotational kinetic energy. The potential energy of a dipole in an electric field is $U=-pE\cos\theta$. If the dipole rotates from an initially unstable (anti-parallel) position $(\theta=\pi)$ to the stable (parallel) alignment $(\theta=0)$, the change in potential energy is:

   $$\Delta U=U(\pi)-U(0) = [pE]-[-pE]=2pE\,.$$

3. Relate potential energy change to rotational kinetic energy:
   The rotational kinetic energy is given by:

   $$\frac{1}{2}I\omega^2\,,$$

   where for a thin ring $I=mR^2$. Equate the energy change to the kinetic energy:

   $$\frac{1}{2}mR^2\omega^2=2pE\,.$$

   Solving for $\omega$:

   $$\omega=\sqrt{\frac{4pE}{mR^2}}\,.$$

4. Substitute $p=\frac{Ql}{2\pi}$ into the expression:

   $$\omega=\sqrt{\frac{4\left(\frac{Ql}{2\pi}\right)E}{mR^2}} =\sqrt{\frac{2QlE}{\pi mR^2}}\,.$$

5. Plug in the given values:
   $Q=100\pi\,\text{C},\; l=0.1\,\text{m},\; E=4\,\text{N/C},\; m=0.1\,\text{kg},\; R=1\,\text{m}$:

   $$\omega = \sqrt{\frac{2\cdot100\pi\cdot0.1\cdot4}{\pi\cdot0.1\cdot1^2}} = \sqrt{\frac{80\pi}{0.1\pi}} = \sqrt{800}\,.$$

   Thus,

   $$\omega=20\sqrt{2}\quad \text{rad/s}\,.$$

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Explanation (Minimal):

• Compute missing charge: $\Delta q = \frac{Ql}{2\pi R}$.
• Effective dipole moment: $p=\Delta q\cdot R=\frac{Ql}{2\pi}$.
• Change in potential energy: $\Delta U = 2pE$.
• Equate $\frac{1}{2}mR^2\omega^2 = 2pE$ and solve for $\omega$.
• Substitute given values to get $\omega=20\sqrt{2}$ rad/s.

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Answer:

$$\boxed{20\sqrt{2} \text{ rad/s}}$$

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Subject, Chapter, and Topic:

• Subject: Physics
• Chapter: Electrostatics & Rotational Motion (from NCERT Class 12; topics include Electric Charges and Fields, and Dynamics of Rotational Motion)
• Topic: Electric Dipoles in a Uniform Electric Field and Energy Conservation in Rotational Motion

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