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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 acceleration
B) Both P and Q do not have any acceleration
C) P has greater acceleration than Q
D) 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,
  & \text{a = }{{\omega }^{2}}r \\
 & \Rightarrow \omega _P = \omega _Q \\
 & \Rightarrow a\propto r \\
 & \therefore a_P > a_Q \\

Additional Information:
Under motion the object may undergo a change in its speed. The rate of change of direction as well as its speed with respect to time is called acceleration. The motion of the object can be linear or circular. Thus, the acceleration involved in linear motion is called linear acceleration. The acceleration involved in circular motion is called angular acceleration.
In uniform circular motion, objects move along a circular path at a constant speed, so acceleration can only point perpendicular to velocity for a change in direction. The acceleration vector must point toward the center to bend the object back on the round path. An external acceleration will bend the direction of the object out and away from the circular path.

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 $m/{sec^2}$.
Angular acceleration is the change in angular velocity divided by time, while tangent acceleration is the change in linear velocity divided by time. People sometimes forget that angular acceleration does not change with radius, but tangential acceleration does.