Question 9: Work, Energy and Power & Circular Motion

Question

A point marked on a ring of radius 2 cm is in contact with a horizontal plane. Now the ring is rolled forward half a revolution along the positive X - direction. Then the angle made by the displacement vector of the point with the X - axis is:

Accepted Answer

Let the plane be the xÃ¢ÂÂaxis and the vertical upward be the yÃ¢ÂÂaxis.
Initial position of the marked point PÃ¢ÂÂ = (0,0).
As the ring (radius R=2Ã¢ÂÂcm) rolls half a turn (ÃÂÃ¢ÂÂrad) without slipping, its center moves horizontally by
$$s = R\cdot\pi = 2\pi\ {m cm}.$$
So the center goes from (0,2) to (2ÃÂ,2).
After a half-turn the point is at the top of the ring, i.e. 2Ã¢ÂÂcm above the center.  Hence its final position is
$$P_f =\bigl(2\pi,\;2+2\bigr) = (2\pi,\,4).$$
The displacement vector is
$$\Delta \mathbf r = (2\pi,\;4)$$
so its angle ÃÂ¸ with the xÃ¢ÂÂaxis satisfies
$$	an	heta = \frac{4}{2\pi} = \frac{2}{\pi}\quad\Longrightarrow\quad      	heta = 	an^{-1}\!\Bigl(\frac{2}{\pi}\Bigr).$$
Answer: Option B.

Answer

$	heta=\boldsymbol{\operatorname { t a n }}^{-\mathbf{1}}\left(\frac{\mathbf{2}}{\mathbf{3} \boldsymbol{\pi}}ight)$

Answer

$\boldsymbol{	heta}=\boldsymbol{\operatorname { t a n }}^{-\mathbf{1}}\left(\frac{\mathbf{2}}{\boldsymbol{\pi}}ight)$

Answer

$	heta=	an ^{-1}\left(\frac{2 \pi}{3}ight)$

Answer

$	heta=\cot ^{-1}\left(\frac{2}{\pi}ight)$

Work, Energy and Power & Circular Motion

A point marked on a ring of radius 2 cm is in contact with a horizontal plane. Now the ring is rolled forward half a revolution along the positive X - direction. Then the angle made by the displacement vector of the point with the X - axis is:

Choose the correct answer: