A circular disc is rotating with angular velocity about an axis that passes through the centre. A particle ‘P’ is kept at a distance of \[2\,m\] from the centre and another particle ‘Q ’at a distance of \[3\,m\] from the centre .Which of these will have higher centripetal acceleration?
Answer
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Hint: In order to answer this question, we must first understand centripetal acceleration. The property of motion of an object traversing a circular path is known as centripetal acceleration. Centripetal acceleration refers to any object travelling in a circle with an acceleration vector pointing towards the circle's centre. We'll also use the equation ${\omega ^2}R$ to solve the problem.
Complete step by step answer:
The acceleration of a body traversing a circular path is known as centripetal acceleration. Since velocity is a vector quantity (it has both a magnitude and a direction), as a body moves in a circular path, its direction changes continuously, causing its velocity to change, resulting in acceleration. The acceleration is radially oriented toward the circle's centre.
The magnitude of the centripetal acceleration ac is equal to the square of the body's speed \[v\] along the curve divided by the distance \[r\] from the circle's centre to the moving body; that is \[{a_c} = {v^2}/r\] in other words, The units for centripetal acceleration are metre per second squared. The force that causes this acceleration is also guided towards the centre of the circle and is named as centripetal force.
If a particle goes in a circle with constant ω then it has a centripetal acceleration given by ${\omega ^2}R$ . since \[\omega \] is constant for both particles. So the acceleration depends on \[R\]
\[\because \;{R_q}\; > \;{R_p}\]
\[\Rightarrow \;{\omega ^2}{R_q}\; > \;{\omega ^2}{R_p}\]
So particle $q$ will have higher centripetal acceleration.
Note: Since we denote centripetal acceleration in the radial direction, the sign of centripetal acceleration is negative. The radial orientation is the direction that radiates outwards from the middle of a circle. We assign the centripetal acceleration a negative sign since it points inwards.
Complete step by step answer:
The acceleration of a body traversing a circular path is known as centripetal acceleration. Since velocity is a vector quantity (it has both a magnitude and a direction), as a body moves in a circular path, its direction changes continuously, causing its velocity to change, resulting in acceleration. The acceleration is radially oriented toward the circle's centre.
The magnitude of the centripetal acceleration ac is equal to the square of the body's speed \[v\] along the curve divided by the distance \[r\] from the circle's centre to the moving body; that is \[{a_c} = {v^2}/r\] in other words, The units for centripetal acceleration are metre per second squared. The force that causes this acceleration is also guided towards the centre of the circle and is named as centripetal force.
If a particle goes in a circle with constant ω then it has a centripetal acceleration given by ${\omega ^2}R$ . since \[\omega \] is constant for both particles. So the acceleration depends on \[R\]
\[\because \;{R_q}\; > \;{R_p}\]
\[\Rightarrow \;{\omega ^2}{R_q}\; > \;{\omega ^2}{R_p}\]
So particle $q$ will have higher centripetal acceleration.
Note: Since we denote centripetal acceleration in the radial direction, the sign of centripetal acceleration is negative. The radial orientation is the direction that radiates outwards from the middle of a circle. We assign the centripetal acceleration a negative sign since it points inwards.
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