Respuesta :
Answer:
N₁ -N₂ = mg [(v₁²-v₂²) / rg + 2]
N₁- N₂ = 2mg
Explanation:
For this problem we apply Newton's second law at the two points
Bottom of the circle
Y Axis
N₁ - W = m a
N₁ = m (a₁ + g)
N₁ = mg (a₁ / g + 1)
Acceleration is centripetal
a₁ = v₁² / r
N₁ = mg (v₁² / rg + 1)
Top of the circle
Y Axis
-N₂ - W = m (-a₂)
N₂ = m (a₂- g)
N₂ = m g (a₂ / g - 1)
a₂ = v₂² / r
N₂ = mg (v₂² / rg -1)
The difference between this normal force is
N₁ -N₂ = mg [v₁² / rg +1 - v₂² / rg +1]
N₁ -N₂ = mg [(v₁²-v₂²) / rg + 2]
In general the speed at the top of the circle is less than the speed at the bottom, as long as you have a system to keep this speed constant, if you keep it constant the result is reduced to
N₁- N₂ = 2mg
The difference between the normal force exerted by the car on a passenger with a mass of m at the top of the loop and the normal force exerted by the car on her at the bottom of the loop is -2mg.
The given parameters;
- mass of the passenger = m
- length of the vertical loop = r
- acceleration due to gravity = g
The normal force on the passenger at the top of vertical loop is calculated as;
[tex]F_n = \frac{mv^2}{r} - mg[/tex]
The normal force on the passenger at the bottom of vertical loop is calculated as;
[tex]F_n = \frac{mv^2}{r} + mg[/tex]
The difference between the two normal forces;
[tex]F_n_{top} - F_n_{down} = \frac{mv^2}{r} -mg -(\frac{mv^2}{r} + mg)\\\\F_n_{top} - F_n_{down} = -2mg[/tex]
Thus, the difference between the normal force exerted by the car on a passenger with a mass of m at the top of the loop and the normal force exerted by the car on her at the bottom of the loop is -2mg.
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