Question

A wheel of radius R rolls on the ground with a uniform velocity "v" the relative acceleration of top most point of the wheel with respect to the bottom most point is___________​

Answer 1

For this we may consider two particles placed at diametrically two opposite points over the wheel.

As the wheel rolls, the two particles on the wheel undergo circular motion about the center of the wheel. So the two particles attain centripetal acceleration since both are in circular motion.

Let the wheel come in such a way that one among the two particles will be at the top most point and the other at the bottom most point. Then the acceleration of the wheel at these two points provide the centripetal acceleration for the particles.

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\put (0,0){\circle{14}}\put(0,7){\vector (0,-1){5}}\put(0,-7){\vector (0,1){5}}\put(0,9){ a_t }\put(0,-10){ a_b }\put(15,0){\vector (0,1){10}}\put(15,0){\vector (1,0){10}}\put(16,10){ Y }\put(27,0){ X }\end{picture}$

The acceleration at the top most point is,

$a_t=-\dfrac {v^2}{R}\ \bf{\hat j}$

The acceleration at the bottom most point is,

$a_b=\dfrac {v^2}{R}\ \bf{\hat j}$

Then the acceleration of the top most point with respect to that of the bottom most point is,

$\boxed {a_t-a_b=-\dfrac {2v^2}{R}\ \bf{\hat j}}$