What is the tension in a rope twirling a 1.5 kg rock in a frictionless circle?

In summary: The apparent weight of a 75.0 kg person traveling at 100 km/hour over the peak of a hill with a radius of curvature = 500m is equal to their mass times the acceleration due to gravity minus the force needed to keep them on their path. At the bottom of the hill, the apparent weight is equal to their mass times the acceleration due to gravity plus the force needed to keep them on their path. In summary, the equations to use are circumference = 2*pi*radius, speed = distance/time, centripetal force = mass*(speed^2)/radius, and force of gravity = mass*9.8m/s^2 down.
  • #1
psychfan29
9
0
1. A 1.5 kg rock is being twirled in a circle on a frictionless surface using a horizontal rope. The radius of the circle is 2.00m and the rope makes 100 revolutions in 1.00 minutes. What is the tension in the rope?

This problem has totally confused me. I don't even know what equation to use that will relate tension to the other givens in the problem!

2. What is the apparent weight of a 75.0 kg person traveling at 100 km/hour:
a) over the peak of a hill with a radius of curvature = to 500m
and
b) at the bottom of a hollow of the same radius

I'm totally lost! HELP!:cry:
 
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  • #2
1) Here are the equations you need:
Circumference = 2*pi*Radius
Speed = Distance / Time
Centripetal Force = Mass * (Speed^2) / Radius
So, if you know the Centripetal Force, what do you think the tension on the rope would be?

2) You should be able to use the equation for centripetal force I gave you here to find the force needed to keep the person on his path. It will be the same magnitude for both at the bottom of the hill and the top, just opposite in direction. When at the top of the hill, some of the force of gravity will be that force keeping the person on the path, and the remaining part will be the apparent weight. At the bottom of the hill, the force of gravity is pushing the person off the path, meaning the normal force pushing the person back on the path (the person's apparent weight) is greater.

Force of Gravity = Mass * 9.8m/s^2 down
 
  • #3
greeniguana00 said:
1) Here are the equations you need:
Circumference = 2*pi*Radius
Speed = Distance / Time
Centripetal Force = Mass * (Speed^2) / Radius
So, if you know the Centripetal Force, what do you think the tension on the rope would be?
I still don't know what equation to use to relate tension to the other values...
 

FAQ: What is the tension in a rope twirling a 1.5 kg rock in a frictionless circle?

What is a frictionless physics problem?

A frictionless physics problem is a scenario where there is no force of friction acting on the objects involved. This means that there is no resistance to the motion of the objects and they can move without any hindrance.

Why are frictionless physics problems used?

Frictionless physics problems are used to simplify real-life scenarios and focus on the fundamental principles of physics. They allow scientists to isolate certain variables and make predictions based on ideal conditions.

How is a frictionless physics problem different from a real-life scenario?

In real-life scenarios, there is always some level of friction present, whether it is due to air resistance, surface roughness, or other factors. Frictionless physics problems ignore these factors to make the problem more manageable and focus on the underlying principles.

What are the limitations of frictionless physics problems?

Frictionless physics problems do not accurately reflect real-life situations, as friction is an important force in our everyday lives. They also do not account for other factors such as air resistance, which can significantly affect the motion of objects.

How can frictionless physics problems be applied in real life?

While frictionless physics problems may not accurately represent real-life scenarios, they can still be used to make predictions and understand the fundamental principles of physics. They can also be used in engineering and design to simplify complex systems and make calculations easier.

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