Asked by natalia
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:
When the car is at the side of the loop (level with the center), find the magnitude of the car's tangential acceleration:
Answers
Answered by
bobpursley
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
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
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