System of Particles and Rotational Motion - Result Question 14
14. A wheel of radius $1 m$ 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
[2002]
(a) $\pi$
(b) $2 \pi$
(c) $\sqrt{2} \pi$
(d) $\sqrt{\pi^{2}+4}$
Show Answer
Answer:
Correct Answer: 14. (d)
Solution:
- (d)
Linear distance moved by wheel in half revolution $=\pi r$. Point $P_1$ after half revolution reaches at $P_2$ vertically $2 m$ above the ground.
$\therefore$ Displacement $P_1 P_2$
$=\sqrt{\pi^{2} r^{2}+2^{2}}=\sqrt{\pi^{2}+4} \quad[\because r=1 m]$
