(╬) See the attachment picture .
(╬) Now, Distance displaced by the point of contact, “AB” =
(╬) And perpendicular distance traversed by the wheel = “2R” .
☞ Therefore, Total distance is
☞ And
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Answers & Comments
(╬) See the attachment picture .
☞ GIVEN:-
(╬) Now, Distance displaced by the point of contact, “AB” =![\tt{({\pi\:R})} \tt{({\pi\:R})}](https://tex.z-dn.net/?f=%5Ctt%7B%28%7B%5Cpi%5C%3AR%7D%29%7D)
(╬) And perpendicular distance traversed by the wheel = “2R” .
☞ Therefore, Total distance is
☞ And![\tt{\boxed{\red{\tan({\theta})\:=\:{\dfrac{BC}{AB}}\:=\:{\dfrac{2}{\pi}}\:}}} \tt{\boxed{\red{\tan({\theta})\:=\:{\dfrac{BC}{AB}}\:=\:{\dfrac{2}{\pi}}\:}}}](https://tex.z-dn.net/?f=%5Ctt%7B%5Cboxed%7B%5Cred%7B%5Ctan%28%7B%5Ctheta%7D%29%5C%3A%3D%5C%3A%7B%5Cdfrac%7BBC%7D%7BAB%7D%7D%5C%3A%3D%5C%3A%7B%5Cdfrac%7B2%7D%7B%5Cpi%7D%7D%5C%3A%7D%7D%7D)
☞ Thus, the point of contact undergoes the displacement of magnitude “
” make an angle “
” with the ground .