The minimum velocity of the bike is C) 18.5 m/s
Explanation:
There are two forces acting on the bike at the top of the loop:
Since the bike is in circular motion, the net force is equal to the centripetal force, so we can write:
[tex]N+mg = m\frac{v^2}{r}[/tex]
where:
m is the mass of the bike
[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity
v is the speed of the bike
r = 35 m is the radius of the loop
Here we are asked to find the minimum speed of the bike, therefore the speed when the bike loses contact with the track, so when the normal reaction is zero:
N = 0
Substituting into the equation and solving for v, we find:
[tex]mg=\frac{v^2}{r}\\v=\sqrt{gr}=\sqrt{(9.8)(35)}=18.5 m/s[/tex]
Which corresponds to option C).
Learn more about circular motion:
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Answer:
C)18.5m/s
Explanation:
At the top of the loop, the centripetal force must equal the weight. Therefore,
Fc=W
[[m[tex]v^{2}[/tex]/r]] = mg
The mass cancels, and does not matter, which simplifies the equation to
[[[tex]v^{2}[/tex]/r]] = g
Solving for v, the minimum velocity of the bike is given by the mathematical expression,
v = [tex]\sqrt{g r}[/tex]
v = [tex]\sqrt{(9.8*35) }[/tex]
= 18.5 m/s
So the bike should have a minimum speed of 18.5 m/s.
