Question

a 1900 kg car starts from rest and drives around a flat 53 - m - diameter circular track. the forward force provided by the cars drive wheels is a constant 1300 n.
part b
what is the direction of the cars acceleration at ( t = 12 s )? give the direction as an angle from the ( r ) - axis.
express your answer in degrees.
( \theta = 15 ^ { circ } )
part c
if the car has rubber tires and the track is concrete, at what time does the car begin to slide out of the circle?
express your answer with the appropriate units.
( t = )

Explanation:

Step1: Calculate the centripetal force

The maximum static frictional force \(f_s=\mu_s N\). Since \(N = mg\) (normal force equals weight for a flat track), and the centripetal force \(F_c=\frac{mv^{2}}{r}\). The tangential acceleration \(a_t=\frac{F}{m}\), where \(F = 1300N\) and \(m = 1900kg\), so \(a_t=\frac{1300}{1900}\approx0.684m/s^{2}\). The velocity \(v=a_t t\). The radius \(r=\frac{53}{2}=26.5m\).

Step2: Set up the equation for the maximum - static - friction condition

The maximum static frictional force provides the necessary centripetal force for circular motion. \(f_s=\mu_s N\) and \(F_c = f_s\). \(N=mg\), \(F_c=\frac{mv^{2}}{r}\), \(v = a_t t\). Substituting \(v\) into the centripetal - force formula, we get \(\mu_s mg=\frac{m(a_t t)^{2}}{r}\). Canceling out \(m\) from both sides of the equation, we have \(\mu_s g=\frac{(a_t t)^{2}}{r}\).

Assume \(\mu_s = 1\) (for rubber tires on concrete, the coefficient of static friction \(\mu_s\approx1\)). Then \(t=\sqrt{\frac{\mu_s gr}{a_t^{2}}}\). Substituting \(g = 9.8m/s^{2}\), \(r = 26.5m\), and \(a_t=\frac{1300}{1900}m/s^{2}\) into the formula:

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Answer:

\(t\approx23.5s\)