Question

A wheel of radius r is rolling on the road.what will be the displacement of a point p in half rotation of wheel

Answer 1

Answer:

R √π² + 4

Explanation:

(pp¹)² = (π²R)² + (2R)²

(pp¹) = √ π²R² + 4R²

        = √R² (π²+4)

        = R √π² + 4

Answer 2

Answer:

The displacement of a point p in half rotation of wheel is $r\sqrt{(\pi^{2} + 4)}$.

Explanation:

Given the radius of a wheel rolling on the road is r.

Assume the point P on the wheel is at the point O, and is in contact with the surface.

When the wheel has completed the half rotation, the point P will be at the top of the wheel.

As shown in the diagram, if we drop a perpendicular from point P, let us say it reaches R.

So, OPR makes a right angled triangle,

where OP is the displacement of the point P.

OR is the half of the perimeter of the circle of radius r, = $\pi r$

and PR is the diameter of the circle of radius r, = $2r$

Applying the Pythagorean theorem,

$OP^{2}= OR^{2}+ RP^{2}$

$\implies OP= \sqrt{(\pi r)^{2}+ (2r)^{2}}$

$\implies OP= \sqrt{r^{2}(\pi^{2} + 4)} =r\sqrt{(\pi^{2} + 4)}$

Thus, the displacement of a point p in half rotation of wheel is $r\sqrt{(\pi^{2} + 4)}$

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