Respuesta :
Answer:
The answers to the question are;
(a) The speed of the rider for your estimated time of 6 s in m/s is 1.91 m/s
(b) The rider's radial acceleration, in m/s² is 2.006 m/s²
(c) The rider's radial acceleration = 8.03 m/s².
Explanation:
To solve the question, we note that
Angular velocity ω is the time to make a given change in the angle of an object per unit time.
Therefore ω = [tex]\frac{\theta}{t} = \frac{S}{r*t} = \frac{v}{r}[/tex]
Where:
θ = Position angle
t = Time
S = Arc length
v = Linear velocity
r = Circle radius
Where the Diameter of the merry-go-round is 12 ft the circumference is given by
Circumference of merry-go-round = π×D = π × 12 ft = 37.7 ft
If it takes 6 seconds make one revolution, then we have
ω₁ [tex]= \frac{37.7}{6*6}[/tex] = 1.047 rad/s
(a) The speed is given by
v = r×ω₁ = 6 ft × 1.047 rad/s = 6.28 ft/s = 1.91 m/s
The speed of the rider for your estimated time of 6 s in m/s = 1.91 m/s
(b) The radial acceleration in m/s² is given by
a[tex]_r[/tex] = ω₁²×r = where r = 6 ft = 1.83 m
(1.047 rad/s)² × 1.83 m= 2.006 m/s²
The rider's radial acceleration, in m/s² = 2.006 m/s²
(c) If the time for one rotation is halved then the angular velocity is doubled and we have
ω₂ = [tex]\frac{\theta}{t}[/tex] = [tex]\frac{2\pi }{\frac{t}{2} }[/tex] = [tex]2*\frac{2\pi }{t}[/tex] = 2×ω₁ = 2×1.047 rad/s = 2.094 rad/s
The radial acceleration in m/s² is then given by
ω₂²× r = (2.094 rad/s)² × 1.83 m = 8.03 m/s²
The rider's radial acceleration = 8.03 m/s²
