A roller coaster car starts from rest at the top of a hill 15 m high and rolls down to ground level. From there it starts into a vertical loop. Exactly at the top of the loop it is 10 m high and the radius of curvature of the track is 4.0 m. Assuming negligible friction and air resistance, what is the magnitude of the normal force acting on a 65 kg passenger at the top of the loop?

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Chenk13 Chenk13 · Answer 1 · 2019-08-06T20:46:04+02:00

Answer:

955.5N

Explanation:

The normal force is given by the difference between the centripetal force and gravity at the top of the loop:

[tex]F_N = F_C - F_G = m\frac{v^{2} }{r} - mg[/tex]

mass m = 65kg

radius of the loop r = 4m

velocity v = ?

g = 9.8 m/s²

To find the centripetal force, you need to find the velocity of the car at the top of the loop.

Use energy conservation:

[tex]E_{tot}=mgh + \frac{1}{2} mv^{2}[/tex]

At the top of the hill:

[tex]E_{tot}= mgh_{hill}[/tex]

At the top of the loop:

[tex]E_{tot}=mgh_{loo}_p +\frac{1}{2} m v^{2}[/tex]

Setting both energies equal and canceling the mass m gives:

[tex]gh_{hill} = gh_{loo}_p + \frac{1}{2} v^{2}[/tex]

Solving for v:

[tex]v^{2} = 2g(h_{hill}-h_{loo}_p)[/tex]

Using v in the first equation:

[tex]F_N = \frac{2mg(h_{hill}-h_{loo}_p)}{r} - mg[/tex]

[tex]F_N = 955.5N[/tex]