Question

A wheel of radius $r$ rolls without slipping with a speed v on a horizontal road. When it is at point A on the road, a small blob of mud separates from the wheel at its highest point and lands at point B on the road.
A. $AB = v\sqrt {\dfrac{r}{g}} $
B. $AB = 2v\sqrt {\dfrac{r}{g}} $
C. $AB = 4v\sqrt {\dfrac{r}{g}} $
D. If $v > \sqrt {4rg} $ , the blob of mud will land on the wheel and not on the road.

Answer 1

Hint:In this question, basically we have to find the distance between the point A and point B. So, first we will check whether the wheel is in rest or in motion at point A that will determine the initial speed of the wheel. And then we will apply the formula of distance in the terms of speed, acceleration (downwards) and the time taken by wheel to reach point B.

Formula used:
$s = ut + \dfrac{1}{2}a{t^2}$
where, $s$ is the distance,$u$ is the initial speed, $a$ is the acceleration, and $t$ is the time taken.

Complete step by step answer:
As per the question, as the body is at point A, that means the body is at rest at its initial time.So, the initial speed is, $u = 0$. And the acceleration is exerted downwards, that means the acceleration is $g$. The blob of mud would have a velocity of $2V$, as the wheel travels from point A to point B, and an initial height is of $2r$.

Thus time of travel $t$ is given by applying the formula of distance in the terms of its velocity, acceleration and time taken:-
The distance will be, $s = 2r$
$\therefore s = ut + \dfrac{1}{2}a{t^2}$
So,
$ \Rightarrow 2r = \dfrac{1}{2}g{t^2}$
Now, we will separate the time to the left hand side:-
$ \Rightarrow t = 2{\sqrt {\dfrac{r}{g}} } $
Therefore, the distance is given by $ = 2vt = 4v\sqrt {\dfrac{r}{g}} $

Hence, the correct option is C.

Note:If a reference rolling object is given, a larger or denser object will roll with the same acceleration. This behaviour is similar to that of an object in free fall or an object sliding down an inclined plane without friction (rather than rolling).