A stuntman drives a car over the top of a hill, the cross section of which can be approximated
by a circular arc of radius 250 m. What is the greatest speed at which he can drive without
the car leaving the road at the top of the hill?
The normal force and the centripetal *force need to equal eachother if the car is to remain grounded, so N=mV^2/r or mg=mV^2/r and then the masses cancel and we have g=V^2/r or v^2=gr.
Well, it seems our stuntman is trying to take his driving skills to new heights, quite literally! To figure out the greatest speed, we need to consider the forces at play. In this case, we're dealing with a circular arc. So buckle up and let's dive into some physics!
To prevent the car from leaving the road, the gravitational force acting on the car needs to be balanced by the centripetal force. Otherwise, we'd have a scene straight out of a cartoon!
The gravitational force pulling the car downwards can be calculated using the mass of the car (which we'll assume is constant) and the acceleration due to gravity. Now, let's define our variables:
m = mass of the car
g = acceleration due to gravity (approximately 9.8 m/s^2)
v = velocity of the car
R = radius of the circular arc (250 m)
The gravitational force pulling the car down is given by F_gravity = m * g.
The centrifugal force keeping the car on the road is given by F_centripetal = m * v^2 / R.
For the car to stay on the road, these forces must be equal. So, we equate them:
m * g = m * v^2 / R
Canceling out the mass (m) on both sides, we have:
g = v^2 / R
Now, we rearrange the equation to find the velocity (v):
v^2 = g * R
v = √(g * R)
Substituting the values for g and R, we get:
v = √(9.8 * 250) ≈ 49.5 m/s
So, our fearless stuntman can drive at a maximum speed of approximately 49.5 m/s (or around 111 mph) without launching his car into orbit! Remember, though, this calculation assumes a perfect world with no friction, air resistance, or other factors that might affect the result. Happy driving, stuntman!
To determine the greatest speed at which the stuntman can drive without the car leaving the road at the top of the hill, we can use the concept of centripetal force.
The force required to keep the car moving in a circular path at the top of the hill is the gravitational force acting on it. At the top of the hill, the gravitational force and the centripetal force need to be equal to maintain circular motion and prevent the car from leaving the road.
The gravitational force can be calculated using the formula F = m * g, where m is the mass of the car and g is the acceleration due to gravity (approximately 9.8 m/s^2).
The centripetal force can be calculated using the formula F = m * (v^2 / r), where m is the mass of the car, v is the velocity, and r is the radius of the circular path. In this case, the radius of the circular arc is 250 m.
Since the gravitational force and the centripetal force need to be equal, we can set up an equation:
m * g = m * (v^2 / r)
The mass of the car cancels out, leaving:
g = v^2 / r
To solve for v, we can rearrange the equation:
v^2 = g * r
v = √(g * r)
Now we can substitute the values:
g = 9.8 m/s^2
r = 250 m
v = √(9.8 * 250)
v = √(2450)
v ≈ 49.5 m/s
Therefore, the greatest speed at which the stuntman can drive without the car leaving the road at the top of the hill is approximately 49.5 m/s.
49.5 m/s
At the top, weight = centripetal force
mg=mv^2/r
solve for v