Answer:
[tex]H=\frac{5}{2}R[/tex]
Explanation:
We use the Energy conservation theorem, in the initial point (h=H, velocity=0, because the car starts from rest) and at the point maximum in the loop (h=2*R ).
[tex]mgH=mg(2R)+\frac{1}{2}mv^2[/tex] 1
but at the top of the loop the car does a circular movement, so the Newtons equation must be fulfilled:
[tex]\Sigma Forces=m*a_{centripetal}[/tex]
[tex]m*g-N=m*v^2/R[/tex]
N is the normal force applied by the loop on the car.
The minimal case occurs when the Normal is equal to zero:
[tex]m*g=m*v^2/R[/tex]
[tex]m*v^2=m*g*R[/tex] 2
We replace 2 in 1, and we solve to find H:
[tex]mgH=mg(2R)+\frac{1}{2}mgR[/tex]
[tex]H=\frac{5}{2}R[/tex]
The minimum height is H in terms of R without losing contact. is given as H = 5R/2.
The sum of kinetic energy and the potential energy of the body is constant. which is the energy of the remains conserved. This is known as energy conversion.
A car starts from rest at point A, a height h above the ground, and rolls freely (no resistance or drag) down a track and around a circular loop of radius R.
Initial velocity will be zero because the car starts from rest.
The maximum point is at 2R.
By the energy conservation theorem, we have
[tex]\rm mgH = mg(2R) + \dfrac{1}{2} mv^2[/tex]... (i)
But at the top of the loop te does a circular movement. Then
[tex]\rm \Sigma F = m a_c\\\\mg - N = m \dfrac{v^2}{R}[/tex]
N is the normal force.
The minimal case occurs when the normal is equal to zero.
[tex]\rm m*g = \dfrac{m*v^2 }{R}\\\\m*v^2 = mgR[/tex] ... (ii)
From equations (i) and (ii), we have
[tex]\rm mgH = mg(2R) + \dfrac{1}{2} mgR\\\\\\H = \dfrac{5}{2}R[/tex]
More about the energy conservation link is given below.
https://brainly.com/question/2137260
