A car of mass 1200 kg is traveling at 55.0 km/hour. It enters a banked turn covered in ice. The road is banked at 20.0 degrees and there is no friction between the road and the cars tires. Use g= 9.80 m/s throughout the problem
Find the radius given the 20 degree angle, assuming the car continues in uniform circular motion around the turn.
Well, this seems like a slippery situation! So, let's put on our thinking caps and solve this problem step by step, shall we?
First things first, we need to find the net force acting on the car in order to keep it in uniform circular motion around the turn. Since there is no friction between the road and the car's tires, the net force will be provided entirely by the horizontal component of the car's weight and the centripetal force.
The horizontal component of the car's weight can be calculated using trigonometry. It is given by: F_horizontal = (mass of car) x (acceleration due to gravity) x sin(angle).
So, F_horizontal = 1200 kg x 9.80 m/s^2 x sin(20 degrees).
Next, we need to determine the centripetal force required for the car to stay in circular motion. The centripetal force is given by: F_centripetal = (mass of car) x (velocity)^2 / (radius of the turn).
We can rearrange this equation to solve for the radius of the turn: radius = (mass of car) x (velocity)^2 / (F_centripetal).
However, since we want to find the radius given the 20-degree angle, we need to relate the centripetal force and the horizontal component of the car's weight.
The centripetal force is equal to the horizontal component of the car's weight, so we can substitute F_centripetal with F_horizontal: radius = (mass of car) x (velocity)^2 / (F_horizontal).
Now, we can substitute the values given into the equation and calculate the radius:
radius = 1200 kg x (55.0 km/hour)^2 / (F_horizontal).
Just remember to convert the velocity from km/hour to m/s. Oh, and don't forget to use appropriate units throughout your calculations!
Now, if you'll excuse me, I'll go juggle some equations to find the radius for you!
To find the radius of the banked turn, we can use the formula for the centripetal force:
F_c = (m * v^2) / r
where F_c is the centripetal force, m is the mass of the car, v is the velocity of the car, and r is the radius of the turn.
In this case, there is no friction between the road and the car's tires, so the centripetal force is provided by the component of the car's weight perpendicular to the road surface. This can be calculated using trigonometry:
F_c = m * g * sin(theta)
where theta is the angle of the banking.
Setting these two equations equal to each other, we can solve for the radius:
(m * v^2) / r = m * g * sin(theta)
Canceling out the mass of the car, we have:
v^2 / r = g * sin(theta)
Now, we can plug in the values:
v = 55.0 km/hour = (55.0 * 1000) / (60 * 60) = 15.3 m/s
g = 9.80 m/s^2
theta = 20.0 degrees = (20.0 * pi) / 180
Plugging these values into the equation, we can solve for r:
(15.3^2) / r = 9.80 * sin((20.0 * pi) / 180)
Simplifying and solving for r, we get:
r = (15.3^2) / (9.80 * sin((20.0 * pi) / 180))
Calculating this value, we find:
r ≈ 59.5 meters
Therefore, the radius of the banked turn is approximately 59.5 meters.
To find the radius of the turn, we can use the concept of centripetal force. In uniform circular motion, the net force acting on an object is directed towards the center of the circular path and is equal to the centripetal force.
In this case, the force of gravity provides the centripetal force as the car moves around the banked turn. The gravitational force can be separated into two components: one perpendicular to the road surface (normal force) and one parallel to the road surface (component of the gravitational force).
Let's break down the problem step by step:
Step 1: Find the normal force (N)
The normal force is the force exerted by the road perpendicular to the surface. In this case, since the road is banked at an angle of 20 degrees, the normal force will also have a component parallel to the road surface.
N * cos(θ) = mg
where N is the normal force, m is the mass of the car (1200 kg), g is the acceleration due to gravity (9.80 m/s²), and θ is the angle of the banked turn (20 degrees).
Step 2: Find the gravitational force component (mg sin(θ))
The component of the gravitational force parallel to the road surface provides the centripetal force.
mg * sin(θ) = (mv²)/r
where v is the velocity of the car (55.0 km/hour = 15.3 m/s) and r is the radius of the turn that we want to find.
Step 3: Combine the equations
Now we can substitute the value of N from step 1 into the equation from step 2:
(mg * cos(θ)) * sin(θ) = (mv²)/r
Simplifying the equation:
mg * cos(θ) * sin(θ) = mv²/r
Step 4: Solve for the radius (r)
Rearrange the equation to isolate r:
r = v² / (g * cos(θ) * sin(θ))
Substitute the given values:
r = (15.3 m/s)² / (9.80 m/s² * cos(20) * sin(20))
Calculating the value:
r ≈ 100.62 meters
Therefore, the radius of the turn is approximately 100.62 meters.
the horizontal component of the normal force (due to the road's banking)
... is equal to the centripetal acceleration
... n sin(20º) = m v^2 / r
the vertical component of the normal force
... is equal to the weight ... otherwise the car would slide vertically
... n cos(20º) = m g
[m g / cos(20º)] sin(20º) = m v^2 / r
r = v^2 / [g tan(20º)]