Question

A particle moves in a 15-cm-radius circle centered at the origin and completes 1.0 rev every 3.0 s. (a) Find the speed of the particle. (b) Find its angular speed o. (c) Write an equation for the x component of the position of the particle as a function of time t, assuming that the particle is on the -x axis at time t 0

Answer 1

a)
`V=0.314 m/s`

b)
`\omega=2.09rad/s`

c)
`x(\theta)=0.15* cos((2\pi)/(3.0) t)`

Step-by-step explanation:

Given:

Radius of the circle, r = 15cm = 0.15m

Time period, T =3.0s

a) The velocity (V) of a particle moving in the circular motion is given as:

`V=(2 \pi r)/(T)`

substituting the given values in the above equation we get

`V=(2* \pi * 0.15m)/(3.0s)`

or

`V=0.314 m/s`

b) Angular speed (
`\omega`) is given as:

`\omega=(2\pi)/(T)`

or

`\omega=(2* \pi)/(3.0s)`

or

`\omega=2.09rad/s`

c) The position of the particle on the x-position is given as:

`x(\theta)=rcos(\theta)` (reffer the attached figure)

now the relation between the Θ and the time T is given as:

`\omega = (2\pi)/(T)=(\theta)/(t)`

or

`\theta= (2\pi)/(T)* t`

or

`\theta= (2\pi)/(3)* t`

substituting the values of r and Θ, we get

`x(\theta)=0.15* cos((2\pi)/(3.0) t)`