A sports car of mass 950kg (including the driver) crosses the rounded top of a hill (radius = 95m) at 22m/s. Determine
a) the normal force exerted by the road on the car,
b) the normal force exerted by the car on the 72kg driver, and
c) the car speed at which the normal force on the driver equals zero.
a) mg-mv^2/r
b) 74(g-v^2/r)
c) when does g=v^2/r >
a) The normal force exerted by the road on the car can be determined by considering the forces acting on the car at the top of the hill. At this point, the only vertical forces acting on the car are the weight of the car and the normal force. Since the car is moving in a circle, there must also be a centripetal force acting towards the center of the circle. In this case, the centripetal force is provided by the difference between the weight and the normal force:
Centripetal Force = Weight - Normal Force
Since the car is not accelerating vertically (it is moving at a constant speed), the vertical net force must be zero:
Net Force (vertical) = 0
This means that the weight is equal in magnitude to the normal force:
Weight = Normal Force
Therefore, the magnitude of the normal force exerted by the road on the car is equal to the weight of the car:
Normal Force = Weight = mass x gravity
Normal Force = 950 kg x 9.8 m/s^2
b) The normal force exerted by the car on the driver is the same as the weight of the driver:
Normal Force on driver = Weight of driver = mass of driver x gravity
Normal Force on driver = 72 kg x 9.8 m/s^2
c) The car speed at which the normal force on the driver equals zero can be determined by considering the forces acting on the driver. At this speed, the centripetal force is equal in magnitude to the weight of the driver:
Centripetal Force = Weight of driver
Since the centripetal force is given by the equation:
Centripetal Force = (mass of driver x car speed^2) / radius
Setting the centripetal force equal to the weight of the driver, we can solve for the car speed:
(mass of driver x car speed^2) / radius = Weight of driver
car speed^2 = (Weight of driver x radius) / mass of driver
car speed = √[(Weight of driver x radius) / mass of driver]
Plugging in the values:
car speed = √[(72 kg x 95 m) / 72 kg]
car speed = √(95 m)
And since I'm a Clown Bot, I'll say that the car speed at which the normal force on the driver equals zero is... when he's not inside the car! Keep the driver safe and sound inside the car at all times!
To solve this problem, we need to consider the forces acting on the car and the driver at the top of the hill.
a) The normal force is the force exerted by the road on the car perpendicular to the surface. At the top of the hill, the car experiences two forces: its weight (mg) directed downwards and the centrifugal force (mv^2/r) directed upwards.
The sum of the vertical forces at the top of the hill is:
Normal force + Weight - Centrifugal force = 0
The weight of the car is given as the mass of the car (including the driver) times the acceleration due to gravity (9.8 m/s^2). The centrifugal force is calculated using the mass of the car and its velocity squared divided by the radius of the hill.
Plugging in the values, we have:
Normal force + (950 kg * 9.8 m/s^2) - [(950 kg * 22 m/s)^2 / 95 m] = 0
Simplifying and solving for the normal force, we get:
Normal force = - (950 * 9.8) + [(950 * 22)^2 / 95] = -8831.7 N
Therefore, the normal force exerted by the road on the car is approximately 8831.7 N, directed downwards.
b) The normal force exerted by the car on the driver is equal in magnitude but opposite in direction to the normal force exerted by the road on the car. Therefore, the normal force exerted by the car on the driver is 8831.7 N, directed upwards.
c) To find the car speed at which the normal force on the driver equals zero, we can set the normal force equal to zero and solve for the velocity.
Normal force = (950 kg * 9.8 m/s^2) - [(950 kg * v^2) / 95 m] = 0
Simplifying and solving for the velocity, we get:
v^2 = (950 * 9.8 * 95) / 950 = 9.8 * 95
Taking the square root of both sides:
v = √(9.8 * 95) ≈ 31.98 m/s
Therefore, the car speed at which the normal force on the driver equals zero is approximately 31.98 m/s.
To solve this problem, we can use the concept of centripetal force and the fact that at the top of the hill, the normal force and gravitational force should add up to provide the centripetal force.
a) The normal force exerted by the road on the car is the force that counters the gravitational force and provides the centripetal force. At the top of the hill, the gravitational force is m * g, where m is the mass of the car (including the driver) and g is the acceleration due to gravity (approximately 9.8 m/s²). The centripetal force is m * v² / r, where v is the velocity of the car and r is the radius of the hill.
So, the normal force exerted by the road on the car is:
Normal force = m * g + m * v² / r
Plugging in the values:
Mass of the car (including the driver) = 950 kg
Gravitational force (Fg) = 950 kg * 9.8 m/s²
Velocity (v) = 22 m/s
Radius (r) = 95 m
Normal force = (950 kg * 9.8 m/s²) + (950 kg * (22 m/s)²) / 95 m
b) The normal force exerted by the car on the driver is equal to the gravitational force acting on the driver. So, it will be equal to the weight of the driver, which is m_driver * g, where m_driver is the mass of the driver.
Mass of the driver = 72 kg
Gravitational force (Fg_driver) = 72 kg * 9.8 m/s²
c) To find the car speed at which the normal force on the driver equals zero, we need to find the velocity at which the gravitational force acting on the driver is exactly countered by the normal force from the car.
Since the normal force on the driver is equal to the gravitational force, we can set up the equation:
Normal force = m_driver * g = 0
So, to solve for the velocity:
m_driver * g = m * v² / r
Plugging in the values:
m_driver = 72 kg
g = 9.8 m/s²
r = 95 m
72 kg * 9.8 m/s² = 950 kg * (v_c² / 95 m)
Solving for v_c:
v_c² = (72 kg * 9.8 m/s² * 95 m) / 950 kg
v_c = √(72 kg * 9.8 m/s² * 95 m) / 950 kg
I hope this explanation helps you understand how to solve the problem!