IDNLearn.com helps you find the answers you need quickly and efficiently. Find reliable solutions to your questions quickly and accurately with help from our dedicated community of experts.
I suppose you meant to say the radius of the curve is 260 m, not mm?
There are 3 forces acting on the car as it makes the turn,
• its weight mg pulling it downward;
• the normal force exerted by the road pointing upward, also with magnitude mg since the car is in equilibrium in the vertical direction; and
• static friction keeping the car from skidding with magnitude µmg (since it's proportional to the normal force), pointing horizontally toward the center of the curve.
By Newton's second law, the net force on the car acting in the horizontal direction is
F = ma => µmg = ma => a = µg
where a is the car's radial acceleration given by
a = v ^2 / R
with v = the car's tangential speed and R = radius of the curve. At the start, the car's radial acceleration is
a = (32 m/s)^2 / (260 m) ≈ 3.94 m/s^2
(a) If µ were reduced by a factor of 2, then the radial acceleration would also be halved:
1/2 a = 1/2 µg
Then the car can have a maximum speed v of
1/2 a = v ^2 / R => v = √(aR/2) = √((3.94 m/s^2) (260 m) / 2) ≈ 22.6 m/s
(b) If µ were increased by a factor of 2, then the acceleration would also get doubled. Then the maximum speed v would be
2a = v ^2 / R => v = √(2aR) = √(2 (3.94 m/s^2) (260 m)) ≈ 45.3 m/s