A rollercoaster car starts at rest on top of a frictionless hill of height 17 m, slides down to ground level, and then into a frictionless loop of radius 5 m.
When the car is at the side of the loop (level with the center), find the magnitude of the car's tangential acceleration:
It fell a distance of 12 m, so it has that energy as velocity.
mg(h)=1/2 mv^2
mg dh/dt= m dv/dt
g dh/dt= dv/dt
now dv/dt is tangential acceleration.
dh/dt is tangential velocity. You can get velocity from
1/2 mv^2=mgh for h=12
To find the magnitude of the car's tangential acceleration at the side of the loop, we can use the law of conservation of energy and centripetal acceleration.
First, let's find the car's velocity at the bottom of the hill before it enters the loop. The height of the hill is 17 m, and since there is no friction, all the potential energy at the top of the hill is converted into kinetic energy at the bottom. Using the law of conservation of energy, we can equate the potential energy to the kinetic energy:
Potential Energy = Kinetic Energy
mgh = (1/2)mv^2
Here, m is the mass of the car, g is the acceleration due to gravity (approximately 9.8 m/s^2), and v is the velocity at the bottom of the hill.
Simplifying the equation, we get:
gh = (1/2)v^2
v^2 = 2gh
v = √(2gh)
Now, let's find the car's acceleration at the side of the loop. The centripetal acceleration is given by the equation:
a = (v^2) / r
Here, r is the radius of the loop, which is 5 m.
Substituting the value of v from the previous equation, we get:
a = (2gh) / r
Substituting the values of g (approximately 9.8 m/s^2), h (17 m), and r (5 m) into the equation, we can calculate the magnitude of the car's tangential acceleration:
a = (2 * 9.8 * 17) / 5
a ≈ 66.672 m/s^2
Therefore, the magnitude of the car's tangential acceleration at the side of the loop is approximately 66.672 m/s^2.