A wheel of radius R rolls on the ground with a uniform velocity v. The relative acceleration of topmost point of the wheel with respect to the bottom most point is

Question

A wheel of radius R rolls on the ground with a uniform velocity v. The relative acceleration of topmost point of the wheel with respect to the bottom most point is (A) (v2/R) (B) (2v2/R) (C) (v2/2R) (D) (4v2/R)

Tardigrade Admin · Accepted Answer

Centripetal acceleration a =( v 2/ R ) towards the centre Thus acceleration of points A and B is shown in the above figure. ∴ Relative acceleration a A B= a A- a B ⇒ a A B=( v 2/ R )-(- v 2/ R )=(2 v 2/ R ) <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/8908e52d3f8373ee156bacf6d26290f8-.png" />

Q. A wheel of radius $R$ rolls on the ground with a uniform velocity $v$ . The relative acceleration of topmost point of the wheel with respect to the bottom most point is

$\frac{v ^{2}}{R}$

$\frac{2 v ^{2}}{R}$

$\frac{v ^{2}}{2 R}$

$\frac{4 v ^{2}}{R}$

Centripetal acceleration $a = \frac{v ^{2}}{R}$ towards the centre
Thus acceleration of points $A$ and $B$ is shown in the above figure.
$∴$ Relative acceleration $a_{A B} = a_{A} - a_{B}$
$\Rightarrow a_{A B} = \frac{v ^{2}}{R} - \frac{- v ^{2}}{R} = \frac{2 v ^{2}}{R}$

Q. A wheel of radius R rolls on the ground with a uniform velocity v. The relative acceleration of topmost point of the wheel with respect to the bottom most point is

Rv2​

R2v2​

2Rv2​

R4v2​