Question

The centre of a wheel rolling on a plane surface moves with a speed v₀​. A particle on the rim of the wheel at the same level as the centre will be moving at speed
A.)Zero
B.)${v_0}$
C.)$2{v_0}$
D.)$\sqrt 2 {v_0}$

Answer 1

Hint- Here the wheel is rolling on the plane surface. Pure rolling is the combination of the two motions, one is pure translation motion of the wheel and another motion is pure rotational motion. Hence this is the combination of the pure translational and pure rotational motion.

Step By Step Answer:
So pure rolling= pure translational motion + pure rotational motion
=in translational motion it is moving with the velocity ${v_0}$ , and in rotational motion it is rotation with the velocity $\omega $,which is, $\omega = \dfrac{{{v_0}}}{R}$, where $R = $radius of the wheel, and this is also the distance of the particle on the rim from the centre.

Since the particle is on the rim of the wheel. So, from here we can get a relation as follows,
${v_0} = R\omega $------equation (1)

So, as there is given in the question that the speed of the centre of the wheel $ = {v_0}$.
Now from the diagram,

At point P,
$v = r\omega $------equation (2)

And from the diagram we see that the value of $r$ can be found as follows,
By Pythagoras theorem,

$r = \sqrt {{R^2} + {R^2}} $
$ \Rightarrow r = \sqrt {2{R^2}} = \sqrt 2 R$
Hence $r = \sqrt 2 R$----equation (3)

Now putting the value of $r$from equation (3) in the equation (1), we get
$v = \sqrt 2 R\omega $

From equation (1) we know that $R\omega = {v_0}$

So finally, $v = \sqrt 2 {v_0}$
Hence the speed of the particle at the rim will be $\sqrt 2 {v_0}$.

Hence the option (D) is the correct answer.

Note- Here the above question is from the topic Kinematics of the circular motion. When a vehicle is moving with some velocity then its velocity will be considered only linear or translational velocity but the wheel of that vehicle will be performing both linear velocity and rotational velocity.