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# A car of mass M = 1300 kg traveling at 45.0 km/hour enters abanked turn...

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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