Answer:
a) 3.97 m/s
b) 0.93 m/s²
c) 0.905
d) 1.905
Step-by-step explanation:
Given,
r = 17 m
t = 27 s
Speed is gotten using
V = 2πr/t
V = 2π*17/27
V = 106.828/27
V = 3.97 m/s
Acceleration would be v²/r
a = 3.97²/17
a = 15.7609/17
a = 0.93 m/s²
Because of the circular motion of the wheel, the net force on the rider at the top of the wheel is downwards. Thus,
Σ F(y) = F(n) - mg = -ma(r) so that
F(n) - mg = -ma(r)
F(n) = mg - ma(r)
F(n) = m[g - a(r)]
F(n) = m(9.8 - 0.93)
F(n) = 8.87m
the ratio of apparent weight to true weight at the top is
W(app) / W = F(n) / mg
8.87*m / 9.8*m
W(app) / W = 0.905
and yet again, as a result of the circular motion, the net force at the bottom of the wheel is upwards.
Σ F(y) = F(n) - mg = ma(r)
F(n) = mg + ma(r)
F(n) = m[g + a(r)]
F(n) = m(9.8 + 0.93)
F(n) = 10.73m
The ratio of apparent weight to true weight at the bottom is
W(app) / W = F(n) / mg
10.73*m / 9.8*m
W(app) / W = 1.095