A car of mass M = 1300 kg traveling at 45.0 km/hour enters a banked turn covered with ice. The road is banked at an angle θ, and there is no friction between the road and the car's tires as shown in (Figure 1) . Use g = 9.80 m/s2 throughout this problem. What is the radius r of the turn if θ = 20.0 ∘ (assuming the car continues in uniform circular motion around the turn)?
What we need to do is the sum of forces in radial direction and vertical direction
\(\sum F_{r}=N \sin \theta=m \frac{v^{2}}{r}\) (1)
\(\sum F_{V}=N \cos \theta=m g\) (2)
then dividing (1)/(2) \(\tan \theta=\frac{v^{2}}{g r}\)
isolating \(r=\frac{v^{2}}{g \tan \theta}=\frac{12.5[\underline{m} s]^{2}}{9.8\left[\frac{m}{s^{2}}\right] \tan 20^{\circ}}=43.8[m]\)
When there is no friction the radius of the turn is \(r=43.8[\mathrm{~m}]\)
given
M = 1300 kg (we do not need this value)
v = 45 km/h
= 45*1000/(60*60) m/s
= 12.5 m/s
theta = 20 degrees
g = 9.8 m/s^2
r = ?
if theta is the angle of banking,
we know, tan(theta) = v^2/(g*r)
r = v^2/(g*tan(theta))
= 12.5^2/(9.8*tan(20))
= 43.8 m
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