Question

A particle travels in a circle of radius 76 cm and completes one revolution in 4.5 s. What is the centripetal acceleration of the particle?

Answer 1

Acceleration,
`a=1.12\ m/s^2`

Step-by-step explanation:

Given that,

Radius of circle, r = 76 cm = 0.76 m

Time period of revolution, t = 4.5 s

To find :

Centripetal acceleration of the particle.

Solution,

Let v is the velocity of particle. On the circular path velocity of a particle is given by :

`v=(2\pi r)/(t)`

The centripetal force on the circular path is given by:

`a=(v^2)/(r)`

`a=(((2\pi r)/(t))^2)/(r)`

`a=(4\pi^2 r^2)/(t^2)`

`a=(4\pi^2 (0.76)^2)/((4.5)^2)`

`a=1.12\ m/s^2`

So, the acceleration of the particle is
`1.12\ m/s^2`.

Answer 2

r = radius of the circle traveled by the particle = 76 cm = 0.76 m

T = time period of revolution for the particle = 4.5 s

w = angular velocity of the particle

angular velocity of the particle is given as

w = 2π/T

inserting the values

w = 2 (3.14)/4.5

w = 1.4 rad/s

a = centripetal acceleration of the particle in the circle

centripetal acceleration is given as

a = r w²

inserting the values

a = (0.76) (1.4)²

a = 1.5 m/s²