Question

1A) A car turning in a circle is acceleratring in the centripetal direction, even if the speed is constant. This centripetal acceleration is the cause of a radially inward directed net force. On a level road this net force is the friction force acting from the road on the tires. You already looked at examples for this.

Find an expression for the speed at which a car can negotiate the turn without any friction in the radial direction (f=0). Calculate that speed (in mph) for a 55 m radius turn that is banked by 9 degrees.

1B) You just should have already derived an expression for the
"frictionless" speed of a car in a banked turn. Now assume that the
coefficient of friction is *u _{s}.* Find an
expression for the maximum speed at which a car can drive through a
turn with radius R and banking angle theta.

Evaluate your answer: test whether in the two limiting cases of "no friction" and "level road" your answer will turn into the two results you already have derived for these cases. Answer with the maximum speed (in mph) for a turn with radius 131 m, banking angle 14 degree, and coefficient of static friction 0.75.

Answer #1

1A) component of normal force(R) on the car by the bank in the radial direction(Fc)

Apply F=ma

along vertical, Rcos(9)-m*g=0; R=mg/cos(9);

along the radias, Rsin(9)=mv^{2}/r ;

mg*sin(9)/cos(9)= mv^{2}/r;

v2=r*g*tan(9)

**v=85.37ms ^{-1}=85.37*3600/(1.6*1000)
= 192.08mph**

1B) Now, friction force= u*N and its along in the direction of radial. Let a- angle,

N-reaction force

Apply F=ma

vertical, Ncos(a)=mg+uNsin(a); N=mg/[cos(a)-usin(a)]

radial, Nsin(a)+u*Ncos(a)=mv^{2}/R ;

So, v^{2}= N*(sin(a)-ucos(a))*R/m =mg/[cos(a)-usin(a)]*
(sin(a)+ucos(a))*R/m

for no friction, u=0

**v=sqrt(gtan(a)*R/m)**
sas canculated

for leveled bank, a=0

**v=sqrt(guR))**

For R=131, a=14; u=0.75

**v=39.72 ms-2
=89.38 mph**

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