91影视

Using the formula for the resultant vector we can find the magnitude of the displacement of the object in 2D and the angle can be calculated using trigonometry.

Formula:

The magnitude of displacement, d is

$d=\sqrt{{\left(螖x\right)}^2+{\left(螖y\right)}^2}$ (1)

When the object is rolling, its center of mass would be moving in such a way that,

$v=r蝇$

Here, r is radius and$蝇$is the angular velocity of rotation. This is the condition for a perfect roll.

When the point P at the circumference rotates for a half rotation, it covers a horizontal distance exactly equal to the distance covered by the center of mass. This distance can be found using the formula for the circumference of the circle.

The horizontal displacement of the wheel $\left(螖x\right)$= its displacement in the half rotation

$\begin{array}{rcl}螖x& =& 蟺脳r\\ & =& 蟺脳0.45\\ & =& 1.4139m\\ & & \end{array}$

At the same time, point P would also move from the bottom to the top. So, the vertical distance traveled would be equal to the diameter of the wheel.

The vertical displacement of the wheel $\left(螖y\right)$= diameter of the wheel

$\begin{array}{rcl}螖y& =& \left(2脳0.45\right)\\ & =& 0.9\text{m}\\ & & \end{array}$

So, using equation (i), the magnitude of the total displacement is,

$\begin{array}{rcl}d& =& \sqrt{{1.414}^2+{0.9}^2}\\ & =& 1.676\\ & 鈮�& 1.68\text{m}\\ & & \end{array}$

Therefore, the magnitude of the displacement is 1.68m.

From the above triangle,

$\begin{array}{rcl}\tan 胃& =& \frac{d}{蟺r}\\ 胃& =& {\tan }^{-1}\left(\frac{0.9}{1.414}\right)\\ & =& 32.48掳\\ & 鈮�& 32.5掳\\ & & \end{array}$

Therefore, the angle of the displacement of P is 32.5^o.