Respuesta :
Critical speed is assumed to be that tangential velocity v where g = v^2/R and g = 9.81 m/sec^2 and R is the radius of rotation. The normal force N = mg - F; and F = mv^2/R = mg when at critical speed, as I defined it. Thus N = mg - mg = 0 at v.
Now, when v = 2v, we have n = mg - m4v^2/R = mg - 4mg so that n/mg = -3 QED. n is the normal force and mg is the gravitational force. The minus sign means the car is going flying unless it is somehow anchored.
The ratio of the Normal force to the gravitational force of the roller coaster at the top of the circle is [tex]\boxed3[/tex].
Further Explanation:
The critical velocity of the roller coaster is the minimum speed at which the roller coaster completes the circular loop without leaving the path.
As the roller coaster reaches at the topmost point of its circular path, at this point, the weight is balanced by the centripetal force acting in the outward direction.
The force balancing at the top most point at the critical velocity is:
[tex]{\text{Centripetal}}\,{\text{force}} = {\text{weight}}[/tex]
[tex]\dfrac{{mv_c^2}}{r} = mg[/tex]
From the above equation, the critical velocity of the roller coaster is:
[tex]{v_c} = \sqrt {rg}[/tex]
Now, if the roller coaster reaches the top point of the circular path with the speed twice of the critical speed, then there will be a normal reaction force experienced by the roller coaster due to the increased centripetal force.
The new velocity of the roller coaster is:
[tex]\begin{aligned}v&=2{v_c}\\&= 2\sqrt {rg}\\\end{aligned}[/tex]
The force balancing at the top most point at the critical velocity is:
[tex]\begin{aligned}{F_C} = {F_N} + {F_{wt}}\hfill\\\frac{{m{v^2}}}{r} = {F_N} + mg\hfill\\\end{aligned}[/tex]
Here, [tex]{F_C}[/tex] is the centripetal force, [tex]{F_N}[/tex] is the normal reaction acting on the roller coaster and [tex]{F_{wt}}[/tex] is the weight of the roller coaster acting in downward direction.
Substitute [tex]2\sqrt {rg}[/tex] for [tex]v[/tex] in above equation.
[tex]\begin{aligned}\frac{{m{{\left( {2\sqrt {rg} }\right)}^2}}}{r} &= {F_N} + mg\\{F_N} &= 4mg - mg\\&= 3mg\\\end{aligned}[/tex]
The ratio of the normal reaction of the roller coaster to the weight of the roller coaster is:
[tex]\begin{aligned}\frac{{{F_N}}}{{{F_{wt}}}}&=\frac{{3mg}}{{mg}}\\&=3\\\end{aligned}[/tex]
The ratio of the Normal force to the gravitational force of the roller coaster at the top of the circle is [tex]\boxed3[/tex].
Learn More:
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Answer Details:
Grade: Senior School
Subject: Physics
Chapter: Circular motion
Keywords: Roller coaster, loop, critical speed, normal reaction, ratio, centripetal force, gravitational force, twice the critical speed, circular loop.
