A car is traveling around a circular banked road without friction which is inclined at 72 degrees which has a vertical height of 25 meters. When the car is traveling at 16 m/s around the road, it is moving at the bottom on the incline (inner radius of the road when viewed from an aerial view). At what speed, in m/s, does the car need to be moving to traveling around the top of the road, which is the outer radius from an aerial view
To find the speed at which the car needs to be moving to travel around the top of the road, we can use the principle of conservation of energy.
When the car is at the bottom of the road, it has maximum gravitational potential energy and no kinetic energy. At the top of the road, it will have maximum kinetic energy and no gravitational potential energy.
We can calculate the speed at the top of the road using the formula for kinetic energy:
KE = (1/2)mv^2
Where KE is the kinetic energy, m is the mass of the car and v is the velocity.
Since we're only interested in the speed, we can ignore the mass of the car and focus on finding the ratio of the kinetic energies at the top and bottom of the road.
The gravitational potential energy at the bottom is given by the formula:
PE = mgh
Where PE is the potential energy, m is the mass of the car, g is the acceleration due to gravity, and h is the vertical height of the road.
Since there is no friction, the total mechanical energy of the system is conserved. Therefore, the sum of the kinetic energy and potential energy at the bottom should be equal to the sum of the kinetic energy and potential energy at the top.
(1/2)mv_bottom^2 + mgh_bottom = (1/2)mv_top^2 + mgh_top
We're interested in finding v_top. Rearranging the equation, we have:
(1/2)v_bottom^2 + gh_bottom = (1/2)v_top^2 + gh_top
Now, let's plug in the given values:
v_bottom = 16 m/s (velocity at the bottom)
h = 25 m (vertical height of the road)
g = 9.8 m/s^2 (acceleration due to gravity)
(1/2)(16)^2 + 9.8(25) = (1/2)v_top^2 + 9.8(0)
Solving for v_top, we get:
128 + 245 = (1/2)v_top^2
373 = (1/2)v_top^2
v_top^2 = 746
Taking the square root of both sides, we find:
v_top = √746
Calculating this value, we get:
v_top ≈ 27.31 m/s
Therefore, the car needs to be moving at approximately 27.31 m/s to travel around the top of the road.