A car traveling 65mph rounds an 100m radius horizontal curve with the tires on the verge of slipping. How fast can this car round a second curve or radius 300m? (assume the same amount of friction between the car's tires and each road surface)
first curve
v=64mph= 29.05m/s
r=100m
second curve
r=300
v=?
To solve this problem would I do this:
v_1*r_1=v_2*r_2
(29.05)(100) = v_2(300)
2905=v_2(300)
v_2=9.68m/s
Is this correct?
If radius increases by a factor of 3, then v^2 increases by 3.
Yes, you are correct in your approach to solving the problem. To find the velocity (v2) at which the car can round the second curve, you can use the conservation of angular momentum, which states that the product of the velocity and radius of a rotating object remains constant.
Using the formula v1*r1 = v2*r2, where v1 is the initial velocity, r1 is the radius of the first curve, v2 is the velocity at the second curve, and r2 is the radius of the second curve, you can solve for v2.
Plugging in the given values, which are v1 = 29.05 m/s (equivalent to 65 mph) and r1 = 100 m for the first curve, and r2 = 300 m for the second curve, you get:
(29.05 m/s)(100 m) = v2(300 m)
2905 = v2(300)
v2 = 2905/300
v2 = 9.68 m/s
So, the car can round the second curve at a speed of approximately 9.68 m/s.