A point P on the circumference of a wheel is | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let the radius of the wheel be $R = 5 \, m$. After half a revolution,the wheel moves forward by a horizontal distance equal to half its circumfere

Vedclass · Accepted Answer

$5\sqrt{\pi^2 + 4} \, m$

Vedclass · Answer

(D) Let the radius of the wheel be $R = 5 \, m$. After half a revolution,the wheel moves forward by a horizontal distance equal to half its circumference,which is $\pi R$. The vertical displacement of point $P$ is equal to the diameter of the wheel,which is $2R$. The total displacement $d$ is the straight-line distance between the initial and final positions of point $P$,given by the Pythagorean theorem: $d = \sqrt{(\pi R)^2 + (2R)^2} = R\sqrt{\pi^2 + 4}$. Substituting $R = 5 \, m$,we get $d = 5\sqrt{\pi^2 + 4} \, m$.

$A$ point $P$ on the circumference of a wheel is initially in contact with the ground. If the radius of the wheel is $5 \, m$,what will be the displacement of point $P$ when the wheel rolls forward by half a revolution?

Similar Questions

When a body moves with a constant speed along a circle,

$A$ particle is moving in $x-y$ plane according to $\vec{r} = b \cos \omega t \hat{i} + b \sin \omega t \hat{j}$, where $\omega$ is a constant. Which of the following statement$(s)$ is/are true?

An object is moving in a circle of radius $100 \, m$ with a constant speed of $31.4 \, m/s$. What is its average speed for one complete revolution?

$A$ particle moves towards east with a velocity of $5\ m/s$. After $10\ s$,its direction changes towards north with the same velocity. The average acceleration of the particle is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module