A rollercoaster car starts at rest on top of a frictionless hill of height 22 m,

Question

A rollercoaster car starts at rest on top of a frictionless hill of height 22 m, slides down to ground level, and then into a frictionless loop of radius 5 m.

a.) When the car is at the side of the loop (level with the center), find the magnitude of the car's:

angular velocity: rad/s
centripetal acceleration: m/s2
tangential acceleration: m/s2

b.) When the car is at the top of the loop, find the magnitude of the car's:

angular velocity: rad/s
centripetal acceleration: m/s2
tangential acceleration: m/s2

Explanation / Answer

At the bottom:
mgh(top) = 1/2mv^2(bottom)
9.8(22) = 1/2*v^2(bottom)
v(bottom) = 20.76 m/s^2

At 5m:
mgh(top) = mgh(half) + 0.5mv^2(half)
9.8(22) = 9.8(5) + 1/2*v^2(half)
v(half) = 18.25 m/s^2

angular velocity = v(half)/r = 18.25/5 = 3.65 rad/s
centripetal acc = (ang vel)^2*r = (3.65)^2 *5 = 66.64 m/s^2

to get tangential acc, I'm going to use kinematics:
v^2(half) = v^2(bottom) + 2a(half)x
18.25^2 = 20.76^2 + 2a(half)*(5)
a(half) = -9.78m/s^2

At 10m:

mgh(top) = mgh(top of loop) + 0.5mv^2(top of loop)
9.8(22) = 9.8(10) + 1/2*v^2(top of loop)
v(topof loop) = 15.33 m/s^2

angular velocity = v(top of loop)/r = 15.33/5 = 3.06rad/s
centripetal acc = (ang vel)^2*r = (3.06)^2 *5 = 46.18 m/s^2

to get tangential acc, I'm going to use kinematics:
v^2(top of loop) = v^2(bottom) + 2a(top of loop)x
15.33^2 = 20.76^2 + 2a(top of loop)*(10)
a(top of loop) = -9.79 m/s^2

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