Answer:
a) [tex]V=0.314 m/s[/tex]
b) [tex]\omega=2.09rad/s[/tex]
c) [tex]x(\theta)=0.15\times cos(\frac{2\pi}{3.0} t)[/tex]
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:
[tex]V=\frac{2 \pi r}{T}[/tex]
substituting the given values in the above equation we get
[tex]V=\frac{2\times \pi \times 0.15m}{3.0s}[/tex]
or
[tex]V=0.314 m/s[/tex]
b) Angular speed ([tex]\omega[/tex]) is given as:
[tex]\omega=\frac{2\pi}{T}[/tex]
or
[tex]\omega=\frac{2\times \pi}{3.0s}[/tex]
or
[tex]\omega=2.09rad/s[/tex]
c) The position of the particle on the x-position is given as:
[tex]x(\theta)=rcos(\theta)[/tex] (reffer the attached figure)
now the relation between the Θ and the time T is given as:
[tex]\omega = \frac{2\pi}{T}=\frac{\theta}{t}[/tex]
or
[tex]\theta= \frac{2\pi}{T}\times t[/tex]
or
[tex]\theta= \frac{2\pi}{3}\times t[/tex]
substituting the values of r and Θ, we get
[tex]x(\theta)=0.15\times cos(\frac{2\pi}{3.0} t)[/tex]
