A wheel of radius $1 \mathrm{~m}$ rolls forward half a revolution on a horizontal ground. The magnitude of…

A wheel of radius $1 \mathrm{~m}$ rolls forward half a revolution on a horizontal ground. The magnitude of the displacement of the point of the wheel initially in contact with the ground is
  1. $\pi$
  2. $2 \pi$
  3. $\sqrt{\pi^{2}+9}$
  4. $\sqrt{\pi^{2}+4}$

Solution



Linear distance moved by wheel in half revolution $=\pi r .$ Point $P_{1}$ after half revolution reaches at $P_{2}$ vertically $2 \mathrm{~m}$ above the ground.
$\therefore$ Displacement $P_{1} P_{2}$ $=\sqrt{\pi^{2} r^{2}+2^{2}}=\sqrt{\pi^{2}+4} \quad[\because r=1 m]$ ^

Asked in: JEE Mains - Rotational Motion - Test 2

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