Question

$\huge\purple {Question:}$

A wheel completes half
a revolution when moving on the road. What is the displacement of the point P when the wheel completes 1/2 a revolution ?​

Answer 1

Given :

➨ No. of revolution = 1/2

(Let radius of wheel be R.)

To Find :

➠ Displacement of point P.

Concept :

➳ Displacement is defined as the shortest distance b/w initial and final position of partical.

➳ It is a vector quantity having both magnitude as well as direction.

➳ It can be positive, negative or zero.

➳ SI unit : m

Diagram :

➛ Please see the attachment for better understanding.

Calculation :

➾ Horizontal distance covered by particle P after half revolution = πR

➾ Vertical distance covered by particle P after half revolution = 2R

$\dashrightarrow\sf\:D(AC)=\sqrt{(AB)^2+(BC)^2}\\ \\ \dashrightarrow\sf\:D=\sqrt{(\pi R)^2+(2R)^2}\\ \\ \dashrightarrow\sf\:D=\sqrt{\pi^2R^2+4R^2}\\ \\ \dashrightarrow\sf\:D=\sqrt{R^2(\pi^2+4)}\\ \\ \dashrightarrow\underline{\boxed{\bf{\purple{D=R\sqrt{\pi^2+4}}}}}\:\gray{\bigstar}$

Answer 2

$\sf{\purple{\underline{\underline{\blue{To\:Find:-}}}}}$

• The Displacement of point, when the wheel completes 1/2 of a revolution .

$\sf{\blue{\underline{\underline{\purple{SOLUTION:-}}}}}$

(╬) See the attachment picture .

• Suppose, Radius of wheel = R

☞ GIVEN:-

• when 1/2 revolution is made by the wheel, the point of contact moves from “A” to “C” and hence undergo the displacement “AC” .

(╬) Now, Distance displaced by the point of contact, “AB” = $\tt{({\pi\:R})}$

(╬) And perpendicular distance traversed by the wheel = “2R” .

☞ Therefore, Total distance is

• $\tt{\green{\:AC\:=\:{\sqrt{{AB}{^2}\:+\:{BC}^{2}}}\:}}$

• $\tt{\red{\implies\:AC\:=\:{\sqrt{({\pi}\:R)^2\:+\:(2R)^2}}\:}}$

• $\tt{\green{\implies\:AC\:=\:R\:{\sqrt{({\pi^{2}}\:+\:4)}}\:}}$

☞ And $\tt{\boxed{\red{\tan({\theta})\:=\:{\dfrac{BC}{AB}}\:=\:{\dfrac{2}{\pi}}\:}}}$

☞ Thus, the point of contact undergoes the displacement of magnitude “ $\tt{\:R{\sqrt{({\pi^{2}}\:+\:4)}}\:}$ ” make an angle “ $\tt{\theta\:=\:{\tan^{-1}({\dfrac{2}{\pi}})}\:}$ ” with the ground .

$\bigstar\:\underline{\boxed{\bf{\red{Required\:Answer\::\:{R{\sqrt{({\pi^{2}}\:+\:4)}}\:}\:}}}}$