Answer with Explanation:
We are given that
Distance,r=0.27 m
Tangential speed=v=0.49 m/s
a.Angular speed ,[tex]\omega=\frac{v}{r}[/tex]
Using the formula
[tex]\omega=\frac{0.49}{0.27}=1.8 rad/s[/tex]
[tex]\omega=\frac{2\pi}{T}[/tex]
[tex]T=\frac{2\pi}{\omega}[/tex]
Time period,[tex]T=\frac{2\pi}{1.8}=3.49 s[/tex]
b.Amplitude,A=Distance of small eraser from the center of a turnable =0.27 m
c.Maximum speed,[tex]V_{max}=0.49 m/s[/tex]
d.Maximum acceleration=[tex]a=r\omega^2=0.27(1.8)^2=0.87 m/s^2[/tex]
Answer:
Explanation:
radius of path, r = 0.27 m
tangential velocity, v = 0.49 m/s
(a) Let the period is T.
Angular speed, ω = v/r
ω = 0.49 / 0.27
ω = 1.81 rad/s
T = 2π/ω
T = ( 2 x 3.14) / 1.81
T = 3.47 s
(b) Amplitude, A = r = 0.27 m
(c) maximum velocity, v = ωA = 1.81 x 0.27 = 0.49 m/s
(d maximum acceleration, a = ω²A = 1.81 x 1.81 x 0.27 = 0.885 m/s²
