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
1 answer:
Answer:
v = 14.92 m/s
Explanation:
First, make a <u>free body diagram</u> and see the <u>forces in the y-direction</u>.
- ∑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 <u>normal force is 0</u> 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
<u>Solve for v</u> 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 <u>14.92 m/s</u>.
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In this case we have,
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