Question

a particle starting from rest moves in a circle of radius 25/pi m with uniformly increasing speed. if it completes 5 revolutions after 5 s of motion the tangential acceleration will be

Answer 1

PHYSICS

ANSWER

Given,

ω0=0rad/s

r=2m

α=4πrad/s2

θ=2πrad

The angular displacement is given by

θ=ω0t+21αt2

2π=214πt

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Answer 2

The tangential acceleration of the particle is 50 m/s^2.

Given:

circle of radius 25/pi

5 revolutions after 5 s

To Find:

tangential acceleration

Solution:

We are aware that the tangential acceleration (at) for an object travelling in a circle is given by:

at = rα

where  is the angular acceleration and r is the circle's radius.

In this instance, the particle is travelling in a circle with a radius of 25 m after beginning at rest. It completes five rotations after moving for five seconds, or one revolution every second.

The circle's diameter, which is equal to 2r = 50 m, is the distance travelled by the particle during one rotation.

So, after moving for a second, the particle's speed is:

v = d/t = 50/1 = 50 m/s

The following method can be used to determine the particle's angular motion ():

ω = v/r

= 50 / (25/π)

= 2π rad/s

The angular acceleration (α) is given by:

α = Δω/Δt

= (ωf - ωi) / t

= (2π*5 - 0) / 5

= 2π rad/s^2

Therefore, the tangential acceleration (at) is:

at = rα

= (25/π) * (2π)

= 50 m/s^2

So, the tangential acceleration of the particle is 50 m/s^2.

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