A rod of length L fixed at the origin at one end is placed along the x-axis at t = 0. It starts rotating with a constant angular acceleration α about the z-axis in the x-y plane in the anti-clockwise as seen from the positive z-axis. The time taken to complete the first revolution is 

  1. \(\sqrt \frac {\pi}{2\alpha}\)
  2. \(\sqrt \frac {\pi}{\alpha}\)
  3. \(\sqrt \frac {2\pi}{\alpha}\)
  4. \(\sqrt \frac {4\pi}{\alpha}\)

Answer (Detailed Solution Below)

Option 4 : \(\sqrt \frac {4\pi}{\alpha}\)
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Detailed Solution

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CONCEPT:

  • Kinematic equations of rotational motion about a fixed axis,

\(\Rightarrow ω = \frac {d θ}{dt}\)

\(\Rightarrow α= \frac {d ω}{dt}\)

\(\Rightarrow α= ω \frac {d ω}{d θ}\)

where ω is the instantaneous angular velocity, α is the instantaneous angular acceleration, θ is the angular displacement, and t is time.

  • Kinematic equations of rotational motion with constant angular acceleration about a fixed axis,

\(\Rightarrow ω = ω _0 + α t\)

\(\Rightarrow θ = ω _0 t + \frac {1}{2} α t^2\)

\(\Rightarrow ω ^2 = ω _0 ^2 + 2 α θ\)

where ω0 is the initial angular velocity, ω is the final angular velocity, α is the angular acceleration, θ is the angular displacement, and t is time.

EXPLANATION:

Given: Length of the rod = L, and angular acceleration = α.

Using the second equation, 

\(\Rightarrow θ = ω _0 t + \frac {1}{2} α t^2\)

\(\Rightarrow 2 \pi = 0 \times t + \frac {1}{2} α t^2\)

\(\Rightarrow T = \sqrt \frac {4\pi}{\alpha}\)

  • Therefore option 4 is correct.  
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