Consider a circular vertical loop-the-loop on a roller coaster. A car coasts without power around the loop. Determine 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. Express your answer in terms of m and the acceleration due to gravity g.

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;

The normal force on the passenger at the top of vertical loop is calculated as;

The normal force on the passenger at the bottom of vertical loop is calculated as;

The difference between the two normal forces;

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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  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

explanation:

v=uf/(u-f)

= 5.-10/(5+10) convex lens has negative focal length

a displacement distance is size of displ