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A point P considered at contact point of a wheel on ground which rolls on ground without slipping then value of displacement of point P when wheel completes half of rotation (If radius of wheel is 1m) : -
A
2 m
B
$\sqrt{\pi^2 + 4}$ m
C
$\pi$ m
D
$\sqrt{\pi^2 + 2}$ m
Detailed Solution
In half a rotation, the wheel rolls forward a distance equal to half its circumference: $\pi R$.
The point P, initially touching the ground, reaches the top of the wheel, i.e. a height 2R above the ground.
So P has a horizontal displacement $PM = \pi R$ and a vertical displacement $QM = 2R$.
Displacement $PQ = \sqrt{(PM)^2 + (QM)^2} = \sqrt{(\pi R)^2 + (2R)^2}$
With R = 1 m: $PQ = \sqrt{\pi^2 + 4}$ m
The point P, initially touching the ground, reaches the top of the wheel, i.e. a height 2R above the ground.
So P has a horizontal displacement $PM = \pi R$ and a vertical displacement $QM = 2R$.
Displacement $PQ = \sqrt{(PM)^2 + (QM)^2} = \sqrt{(\pi R)^2 + (2R)^2}$
With R = 1 m: $PQ = \sqrt{\pi^2 + 4}$ m