A curve of radius 60m is banked so that a car traveling with uniform speed 70km/hr can round the curve without relying on friction to keep it from slipping to its left or right. The acceleration of gravity is 9.8m/s^2. What the the Angle of the curve?
im using a=v^2/r, but when I get the acceleration, I don't know what to do no more.. Can anyone help me on this one? Thanks a bunch!
You do not have to rely of friction to keep the car from slipping if the road is banked at an angle such that the force applied by the road to the tires is perpendicular to the surface of the road. The horizontal component of that force, in that situation, equals the centripetal acceleration M V^2/R and the vertical component balances the weight, M g.
By drawing yourself a figure, you should be able to convince yourself that the "bank angle" of the curve is
theta = arctan [(MV^2/R)/Mg]
= arctan [V^2/(R g)]
Make sure you express V in m/s when using the formula.
To find the angle of the curve, you can use the formula:
θ = arctan(V^2 / (R * g))
where:
- θ is the angle of the curve
- V is the velocity of the car in meters per second (m/s)
- R is the radius of the curve in meters (m)
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
In this case, the radius of the curve is given as 60 meters and the velocity of the car is given as 70 km/hr.
First, let's convert the velocity from km/hr to m/s.
Using the conversion factor 1 km/hr = 1000 m/3600 s, we can convert 70 km/hr to m/s:
70 km/hr * (1000 m/3600 s) = 19.44 m/s
Now, substitute the values into the formula:
θ = arctan(19.44^2 / (60 * 9.8))
θ = arctan(378.05 / 588)
Using a calculator, find the value of arctan(0.6432) to get the angle of the curve.
θ ≈ 32.49°
Therefore, the angle of the curve is approximately 32.49°.