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A rollercoaster loop has a radius of 22.7 m. what is the minimum speed the coaster must have at the top of the loop to not fall off the track? (unit = m/s)

Sagot :

Supernova

Answer:

v = 14.92 m/s

Explanation:

First, make a free body diagram and see the forces in the y-direction.

  • ∑F_y = F_N - F_g

Use Newton's 2nd Law F = ma to replace ∑F_y with m * a_y.

  • m * a_y = F_N - F_g

The acceleration in the y-direction is the centripetal acceleration, a_c = v^2/r.

  • m * v^2/r = F_N - mg

The normal force is 0 because this is where the rollercoaster is not falling off the track yet not touching the track.

  • m * v^2/r = - mg

The masses cancel out.

  • v^2/r = -g

Solve for v to find the speed of the rollercoaster at the top of the loop.

  • v^2 = -(-9.81) * r
  • v = √(9.81 * 22.7)
  • v = 14.9227

The minimum speed the coaster must have at the top of the loop to not fall off the track is 14.92 m/s.

View image Supernova
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