Amusement Park Centripetal Force Around A Loop

In summary, the amusement park ride called The Roundup involves passengers standing inside an 18.0 m-diameter rotating ring that tilts into a vertical plane. The ring rotates once every 4.90 s and the rider's mass is 58.0 kg. To find the force at the top, one must consider the weight and radial acceleration acting on the rider. The equation for force at the top is (m*v^2)/r, where m is the mass, v is the velocity, and r is the radius. At the bottom, the forces are opposing and the normal force must also be taken into account. To find the longest rotation period of the wheel that will prevent riders from falling off at the top, one
  • #1
GoSS190
20
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In an amusement park ride called The Roundup, passengers stand inside a 18.0 m-diameter rotating ring. After the ring has acquired sufficient speed, it tilts into a vertical plane. Suppose the ring rotates once every 4.90 s. and the rider's mass is 58.0 kg.

A.) With how much force does the ring push on her at the top of the ride?

B.) With how much force does the ring push on her at the bottom of the ride?

C.) What is the longest rotation period of the wheel that will prevent the riders from falling off at the top?

If anyone could help me with these questions that would be great. Thanks



I tried this using the equations to find v = (2pi(r)) / T

but then i realized that the velocity is different at the top than at the bottom. I am stumped as to how to find the velocity then find the force.

I think the equation for force at the top is (m(vtop)2) / R

Can anyone help me out with this though please
 
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  • #2
GoSS190 said:
In an amusement park ride called The Roundup, passengers stand inside a 18.0 m-diameter rotating ring. After the ring has acquired sufficient speed, it tilts into a vertical plane. Suppose the ring rotates once every 4.90 s. and the rider's mass is 58.0 kg.

A.) With how much force does the ring push on her at the top of the ride?

B.) With how much force does the ring push on her at the bottom of the ride?

C.) What is the longest rotation period of the wheel that will prevent the riders from falling off at the top?

If anyone could help me with these questions that would be great. Thanks

I tried this using the equations to find v = (2pi(r)) / T

but then i realized that the velocity is different at the top than at the bottom. I am stumped as to how to find the velocity then find the force.

I think the equation for force at the top is (m(vtop)2) / R

Can anyone help me out with this though please

No. That's not what's going on.

Draw a force diagram. You have 2 forces acting on the 58 kg person.

There's weight - which always acts down and there is the radial acceleration from the rotation of the ride. So there are two forces. What is the difference between looking at the forces between the top and the bottom then?
 
  • #3
The force at the top would be adding to each other while at the bottom they would be opposing. But how do u find the normal force?
 

FAQ: Amusement Park Centripetal Force Around A Loop

What is centripetal force in the context of an amusement park loop?

Centripetal force is the force that keeps an object moving in a curved path. In the case of an amusement park loop, it is the force that keeps the rider moving in a circular path around the loop.

How is centripetal force generated in an amusement park loop?

Centripetal force is generated by the combination of the rider's inertia, which wants to keep them moving in a straight line, and the normal force from the track, which pushes the rider towards the center of the loop.

What factors affect the centripetal force experienced by a rider in an amusement park loop?

The factors that affect centripetal force include the speed of the rider, the mass of the rider, and the radius of the loop. The greater the speed and mass of the rider, and the smaller the radius of the loop, the greater the centripetal force will be.

What happens if there is not enough centripetal force in an amusement park loop?

If there is not enough centripetal force, the rider will not be able to complete the loop and will either fall off the track or continue moving in a straight line tangent to the loop. This is why it is important for the ride to have enough speed and a properly sized loop to generate enough centripetal force.

How can we calculate the centripetal force experienced by a rider in an amusement park loop?

The centripetal force can be calculated using the equation Fc = mv^2/r, where Fc is the centripetal force, m is the mass of the rider, v is the speed, and r is the radius of the loop. This equation can also be rearranged to solve for any of these variables if the others are known.

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