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

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

Answer 2

Answer:

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

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 :

$v=\dfrac{2\pi r}{t}$

The centripetal force on the circular path is given by:

$a=\dfrac{v^2}{r}$

$a=\dfrac{(\dfrac{2\pi r}{t})^2}{r}$

$a=\dfrac{4\pi^2 r^2}{t^2}$

$a=\dfrac{4\pi^2 (0.76)^2}{(4.5)^2}$

$a=1.12\ m/s^2$

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