A 300 kg rollercoaster car starts at rest on top of a frictionless hill of heigh

Question

A 300 kg rollercoaster car starts at rest on top of a frictionless hill of height 13.1 m, slides down to ground level, and then into a frictionless loop of radius 5 m. Which of the listed techniques would be practical/useful to analyze this situation? When the car is at the side of the loop (level with the center) on the way up; Find the magnitudes of the following quantities for the car: angular velocity: centripetal acceleration: tangential acceleration: normal force from the track: Also, find F_net on the car at this moment When the car is at the top of the loop: Find the magnitudes of the following quantities for the car: angular velocity: centripetal acceleration: tangential acceleration: normal force from the track: Also, find F_net on the car at this moment:

Explanation / Answer

A) Energy method

B) v = r*w

w= v/r

apply law of conservation of energy

0.5*m*v^2 + m*g*R = m*g*h

v = 12.6 m/s

then w = 12.6/5 =2.52 rad/s

centripetal accelaration a = r*w^2 = 5*2.52*2.52 = 31.752 m/s^2

tangential accelaration = a = dv/dt = 12.6/(T/4)

T = 2*pi*r/V = 2*3.142*5/12.6 =2.5 s

a = 4*12.6/2.5 = 20.16 m/s^2

normal force from the track N= m*g = 300*9.81 = 2943 N

Fnet = m*a = m*v^2/r = 300*12.6*12.6/5 = 9525.6 N and 90 degree from the velocity

C)

m*g*h = 0.5*m*v^2 + m*g*2R

v = 7.8 m/s

angular velocity w = v/r = 7.8/5 = 1.55 rad/s

centripetal accelaration a = v^2/r = 7.8*7.8/5 = 12.168 m/s^2

tangential accelation a = v/(T/4) = 4*v/T = 4*7.8/2.5 = 12.48 m/s^2

Normal force N

mg-N = m*v^2/r = 300*7.8*7.8/5 = 3650.4

N = (300*9.81)-3650.4 = 707.4 N in the upward direction

Fnet = m*v^2 /r = (300*7.8*7.8/5) = 3650.4 N and 90 degrees from the velocity

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