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PhysicsRotationRotational KinematicsEasy2 minPYQ_2021

PhysicsEasynumerical

The centre of a wheel rolling on a plane surface moves with a speed $v_{0} .$ A particle on the rim of the wheel at the same level as the centre will be moving at a speed $\sqrt{x} v_{0} .$ Then the value of $x$ is $.$

Question diagram: The centre of a wheel rolling on a plane surface moves with

Answer:

2.00

Solution:

For no slipping, $v_{0} = ω R$
Now, $v_{A} = v_{B} = \sqrt{v_{0}^{2} + (ω R)^{2}}$
$= \sqrt{2} v_{0}$

$\Rightarrow x = 2$

Stream:JEESubject:PhysicsTopic:RotationSubtopic:Rotational Kinematics

⏱ 2mℹ️ Source: PYQ_2021

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