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# Two particles P and Q are located at distances $r_P$​ and $r_Q$​ respectively from the axis of a rotating disc such that $r_P$​ > $r_Q$​.A) Both P and Q have the same accelerationB) Both P and Q do not have any accelerationC) P has greater acceleration than QD) Q has greater acceleration than P

Last updated date: 20th Jun 2024
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Hint: Radial acceleration is directly proportional to the product of the angular momentum and the radius of the curved path.

Complete step by step solution:
Angular speed $\omega$ does not change with radius, but linear speed v does.
As the objects are not moving and $r_P$ is greater than $r_Q$ so,
\begin{align} & \text{a = }{{\omega }^{2}}r \\ & \Rightarrow \omega _P = \omega _Q \\ & \Rightarrow a\propto r \\ & \therefore a_P > a_Q \\ \end{align}

Note: In circular motion the acceleration is toward the center of a curved path and perpendicular to the velocity of the object. An object has to change its direction and not its movement along a circular path. Also called radial acceleration. SI unit is $m/{sec^2}$.