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>.
You might be interested in
Answer:
11.4 m/s
Explanation:
The expression for the Centripetal acceleration is :
Where, a is the accleration
v is the velocity around circumference of circle
R is radius of circle
In the given question,
a = g = Acceleration due to gravity as the car is at top =
v = ?
R = 13.2 m
So,
<u>v = 11.4 m/s</u>
Answer:
The answer to your question is: F = 0.4375 N. The force will be 16 times lower than with the first conditions.
Explanation:
Data
F = 7 N
F = ? if the masses is quartered
Formula
Process
Normal conditions F = Km₁m₂/r² = 7
When masses quartered F = K(m₁/4)(m₂/4)/r² = ?
F = K(m₁m₂/16)/r²
F = K(m₁m₂/16r² = 7/16 = 0.4375 N