At what minimum speed must a roller coaster be traveling when upside down at the top of a circle so that the passengers will not fall out? Assume a radius of curvature of 8.47 m.
(in m/s)
Well, imagine this: if the roller coaster goes too slow, the passengers might start falling out of the seats, which is not a pretty sight. Nobody wants to see people flying around like a bunch of confused birds. So, to prevent this catastrophe, we need to calculate the minimum speed required at the top of the loop-de-loop.
Now, gravity is always trying to pull the passengers downward, so we need to make sure the centrifugal force pushing them outward is greater than or equal to the gravitational force pulling them downward.
To find this minimum speed, we can equate the two forces:
mv^2/r = mg
Here, m is the mass of a passenger, v is the speed, r is the radius of the circle, and g is the acceleration due to gravity.
Simplifying a bit, we get:
v^2 = rg
Substituting in the values, with a radius of 8.47 m and a gravitational acceleration of 9.8 m/s^2, we can solve for v:
v^2 = 8.47 m * 9.8 m/s^2
v^2 = 82.966 m^2/s^2
And taking the square root of both sides, we find:
v = √82.966 m/s
So, my dear friend, the minimum speed the roller coaster must be traveling at the top of the circle is approximately 9.11 m/s. And if you ask me, that's pretty fast for clowns like me. Hold on tight! 🎢😄
To determine the minimum speed a roller coaster must be traveling when upside down at the top of a circle, we can use the concept of centripetal force.
At the top of the loop, the force exerted by the roller coaster on the passengers (centripetal force) must be greater than or equal to the gravitational force pulling the passengers downward to prevent them from falling out.
The centripetal force is given by the equation:
Fc = mv^2 / r
Where Fc is the centripetal force, m is the mass of the passenger, v is the velocity of the roller coaster, and r is the radius of curvature.
At the top of the loop, the gravitational force pulling the passengers downward is given by:
Fg = mg
Where Fg is the gravitational force and g is the acceleration due to gravity.
Setting these two forces equal to each other, we have:
mv^2 / r = mg
To find the minimum speed, we can assume that the centripetal force is just equal to the gravitational force (Fc = Fg).
So the equation becomes:
mv^2 / r = mg
Canceling out the mass (m) on both sides of the equation, we have:
v^2 / r = g
Solving for the minimum velocity (v), we can rearrange the equation as follows:
v = √(rg)
Plugging in the values for the radius of curvature (r = 8.47 m) and the acceleration due to gravity (g = 9.8 m/s^2), we can calculate the minimum speed required:
v = √(8.47 * 9.8)
v = √(82.966)
v ≈ 9.11 m/s
So, the minimum speed at which the roller coaster must be traveling when upside down at the top of the circle is approximately 9.11 m/s.
To determine the minimum speed required for passengers on a roller coaster to not fall out when upside down at the top of a circle, we can use the concept of centripetal force. At the top of the circle, the centripetal force acting on the passengers must be equal to or greater than the force due to gravity pulling them downward.
The centripetal force (Fc) is given by the formula:
Fc = m * v² / r
Where:
m = mass of the passenger
v = velocity of the roller coaster
r = radius of curvature
In this case, we need to solve for the minimum velocity (v) required. We can consider the gravitational force (Fg) acting on the passengers, which is simply the weight:
Fg = m * g
Where:
m = mass of the passenger
g = acceleration due to gravity (9.8 m/s²)
At the top of the circle, the centripetal force (Fc) should be equal to the gravitational force (Fg). So we can set up the equation:
Fc = Fg
m * v² / r = m * g
Simplifying the equation:
v² = r * g
v = √(r * g)
Now we can plug in the values:
r = 8.47 m
g = 9.8 m/s²
v = √(8.47 * 9.8)
v ≈ 9.8 m/s
Therefore, the minimum speed the roller coaster must be traveling at the top of the circle, so the passengers will not fall out, is approximately 9.8 m/s.