Answer:
15m
Explanation:
Given that a roller coaster starts from rest at A, rolls down the track to B, describes a circular loop of 12-m diameter, and travels up and down past point E. Determine the range of values of h for which the roller coaster will not leave the track at D or E. Assume no energy loss due to friction
Solution
At point A
The maximum potential energy = maximum K.E
At point A, the total energy = maximum P.E.
Down the track to point B, the P.E will be converted to maximum K.E.
Hence,
Mgh = 1/2mv^2
Also, the total energy at the roller coaster will be P.E + K.E
I.e mg2r + 1/2mv^2
Where 2r = height of the loop = diameter of the loop.
Since the energy is always conserved, hence
Mgh = mg2r + 1/2mv^2
Let also consider the centripetal acceleration to keep the object in the circle.
F = mV^2 / r = mg
Mass will cancel out
U^2 = rg
Substitute that in the last equation
Mgh = mg2r + 1/2mgr
mgh = mg ( 2r + 1/2r )
Mg will cancel out
h = 2.5r
Where r = 12/2 = 6
h = 2.5 × 6
h = 15m
Therefore, the values of h for which the roller coaster will not leave the track at D or E is 15m.
