Answer:
Explanation:
From the given information:
radius = 15 m
Time T = 23 s
a) Speed (v) = [tex]\dfrac{2 \pi r}{T}[/tex]
[tex]v = \dfrac{2\times \pi \times 15}{23}[/tex]
v = 4.10 m/s
b) The magnitude of the acceleration is:
[tex]a = \dfrac{v^2}{r} \\ \\ a = \dfrac{(4.10)^2}{15}[/tex]
a = 1.12 m/s²
c) True weight = mg
Apparent weight = normal force
From the top;
the normal force = upward direction,
weight is downward as well as the acceleration.
true weight - normal force = ma
apparent weight =mg - ma
[tex]\dfrac{apparent \ weight}{true \ weight} = \dfrac{(mg - ma)}{(mg)}[/tex]
[tex]=1- \dfrac{1.12}{9.8}[/tex]
= 0.886 m/s²
d)
From the bottom;
acceleration is upward, so:
apparent weight - true weight = ma
apparent weight = true weight + ma
[tex]\dfrac{apparent \ weight }{true \ weight} =\dfrac{ mg + ma}{mg}[/tex]
[tex]\dfrac{apparent \ weight }{true \ weight} =1+ \dfrac{a}{g} \\ \\ = 1 + \dfrac{1.12}{9.8}[/tex]
[tex]= 1 + \dfrac{1.12}{9.8} \\ \\[/tex]
= 1.114 m/s²
