Question

A point P consider at contact point of a wheel on ground which rolls on ground without slipping then value of displacement of point P when wheel completes half of rotation (If radius of wheel is 1m) : 

• A. 2m
• B. \(\sqrt{\pi^2+4}\) m
• C. \(\pi\) m
• D. \(\sqrt{π^2+2}\) m

Answer 1

The Correct Option is B

To find the displacement of a point P on the circumference of a wheel when the wheel completes half a rotation and rolls without slipping, we need to consider both the rotation and the translation of the wheel on the ground.

The wheel has a radius \( r = 1 \) m. When it completes half a rotation, the arc length covered by point P due to rotation, which is half of the wheel’s circumference, is given by:

L_{\text{arc}} = \frac{1}{2} \times 2\pi r = \pi \, \text{m}

As the wheel rolls without slipping, the center of the wheel moves a linear distance equal to the arc length traveled by the wheel. So, the center of the wheel translates horizontally by:

d_{\text{center}} = \pi \, \text{m}

However, the actual path traveled by point P is not just horizontal or along the arc. The total displacement \(|d_{\text{total}}|\), is the straight-line distance from the initial position to the final position of point P.

Initially, point P is at the contact point on the ground. After half rotation and translation by \(\pi\) m, point P is again on the ground but on the opposite side of the circle. Its vertical change from the starting contact point to the highest point and down again to the contact point contributes fully to the diameter of the wheel:

d_{\text{vertical}} = 2 \times r = 2 \, \text{m}

Using these, we calculate the straight-line displacement using Pythagoras' theorem, where the horizontal movement is \(\pi\) m and vertical movement is 2 m:

|d_{\text{total}}| = \sqrt{(\pi)^2 + 2^2} = \sqrt{\pi^2 + 4} \, \text{m}

Therefore, the correct displacement of point P when the wheel completes half a rotation is:

\(\sqrt{\pi^2+4}\) m

Thus, the correct answer is: \(\sqrt{\pi^2+4}\) m