Question

A wheel of radius 1 metre rolls forward half a Revolution on a horizontal ground find the magnitude of the displacement of the point of the wheel initially in contact with the ground. PLzz urg ent be fast

Answer 1

Given, the radius of the wheel is 1m. so,its circumference is 2π m
when it rotates half, the initial point of contact is vertically above the final point of contact.thus,the three points form a right triangle.

by pythagoras theorem, we solve it to get the magnitude of displacement,which is the hypotenuse.
thus, d=√4+π²




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Answer 2

Given that,

A wheel of radius 1 metre rolls forward and completes half of a revolution on a horizontal ground . Now,We have to find the displacement from the point of wheel that was initially in contact with ground.

Here we go,

We can find three points that will be form a right angled triangle.
First point will be the point of wheel that was in contact with ground another point will be the same point of wheel that will be in air and finally third point will be the point that we shall get by drawing a perpendicular from second point to earth.
C
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|
|
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|______________ A
B

A is the point of wheel that was initially in contact with Earth,C is its final point.

Distance from A to B = Half of the circumference

= 2πr ÷ 2

= πr

= π × 1 m

= π m

Distance from B to C will be the diameter of circle = 2 m.

So,

By Pythagoras Theroem,

=> AC = √( AB² + BC² )

=> AC = √[ ( 2m )² + ( π m )² ]

=> AC = √( 4 m² + π² m² )

=> AC = √( 4 + π² )m

Hence,the displacement is √( 4 + π² )m.