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JEE Main 2020 (9 Jan, Shift 1) · Rotational Kinetic Energy, Work & Power

Rotational Kinetic Energy, Work & Power question from JEE Main 2020 (9 Jan, Shift 1)

Q43JEE Main 2020 · 9 JAN · SHIFT 1Medium

Rotational Kinetic Energy, Work & Power

One end of a straight uniform 1m long bar is pivoted on horizontal table. It is released from rest when it makes an angle 30° from the horizontal (see figure). Its angular speed when it hits the table is given as n\sqrt{n} s-1, where n is an integer. The value of n is _________.

Figure

Numerical answer type. Work it out, then open the answer.

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Answer: 15

  1. Centre of mass falls L2sin⁡30∘=12×12=0.25\frac{L}{2}\sin 30^\circ = \frac{1}{2}\times\frac{1}{2} = 0.25 m.
  2. Mg(0.25)=12⋅ML23ω2Mg(0.25) = \frac{1}{2}\cdot\frac{ML^2}{3}\omega^2 with L=1L = 1 m
  3. ω2=6g(0.25)=1.5×10=15⇒ω=15\omega^2 = 6g(0.25) = 1.5\times 10 = 15 \Rightarrow \omega = \sqrt{15} s−1^{-1}
  4. So n=15n = 15 (taking g=10g = 10 m/s2^2).
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