A racecar driver speeds around a circular track of radius 100 m at a velocity of 50 km/hr. The coefficient of friction us between his tires and the road is 0.20. It begins to rain, reducing us to 0.15. What must he change his velocity to in order to maintain his path without skidding off of the track? (use units of km/hr.).
When on the verge of skidding
M V^2/R = M g * (mu,s)
mu,s is the static coeffiecent of friction, 0.15 in this case. Cancel the M's and solve for V. It wil be in m/s if you use g = 9.8 m/s^2, so you will have to convert that V
to km/h
To solve this problem, we can use the equation:
MV^2/R = Mg * μs
Where:
M is the mass of the racecar driver,
V is the velocity of the racecar driver,
R is the radius of the circular track,
Mg is the weight of the racecar driver (M multiplied by the acceleration due to gravity), and
μs is the static coefficient of friction.
In the given problem, the coefficient of friction changes from μs = 0.20 to μs = 0.15 due to rain. We need to find the new velocity that the racecar driver needs to maintain to prevent skidding.
First, we cancel out the mass (M) on both sides of the equation since it appears on both sides. The equation becomes:
V^2/R = g * μs
Now, we can substitute the known values into the equation:
V^2/100 = 9.8 * 0.15
Simplifying further:
V^2 = 100 * 9.8 * 0.15
V^2 = 147
Taking the square root of both sides:
V ≈ √147
V ≈ 12.12 m/s (meters per second)
However, we need to convert this velocity to km/hr, as the question requires the answer in that unit.
To convert m/s to km/hr, we use the conversion factor: 1 km = 1000 m and 1 hr = 3600 sec. So, 1 m/s = (1 km/1000 m) * (3600 sec/1 hr).
Substituting the value of V:
V (km/hr) = 12.12 m/s * (1 km/1000 m) * (3600 sec/1 hr)
Now, perform the calculations:
V (km/hr) = 12.12 * (1/1000) * 3600
V (km/hr) ≈ 43.6 km/hr
Therefore, the racecar driver must change their velocity to approximately 43.6 km/hr to maintain their path without skidding off the track.