Question

A particle completes 4 revolution per second on a circular path of a radius 0.25 m. Calculate the angular velocity and the centripetal acceleration of the particle

Answer 1

Answer:

Explanation:

consider Frequancy f=4

r=0.25m

then angle subtended in 4 revolutions

theta=4×2pi=8pi rad

angular velocity=theta/t=8pi×frequancy

angular velocity=32pirad/s

centripetal accalration=(angular velocity)^2×radius

centripetal accalration=(32×22/7)^2×0.25

Answer 2

The angular velocity and the centripetal acceleration of the particle is $25.12 rad/s$ and $157.75 m/s^{2}$ respectively.

Given,

Radius of the circular path=0.25 m

Frequency(revolutions per minute or per second) of the motion=4 revolutions per second.

To find,

the angular velocity and the centripetal acceleration of the particle.

Solution:

• The angular velocity(speed), ω of a particle with a time period T is represented by the following expression:
• $w=\frac{2\pi }{T}$.
• The time period, T and the frequency(revolutions per minute or per second), f are related as:
• $f=\frac{1}{T}$.

• Thus the angular velocity of a particle can also be given as:
• $w=2\pi f$.
• The acceleration due to change in the direction of velocity is centripetal acceleration.
• It is represented by the following expression:
• $a=Rw^{2}$.
• where, R-radius of the circle.

The angular velocity of the particle will be:

$w=2\pi f\\w=2X3.14X4\\w=25.12 rad/s.$

The centripetal acceleration of the particle will be:

$a=Rw^{2}\\a=0.25X25.12^{2} \\a=157.7536\\a=157.75 m/s^{2} .$

Hence, the angular velocity and the centripetal acceleration of the particle is $25.12 rad/s$ and $157.75 m/s^{2}$ respectively.

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