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

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SUMMARY

The discussion centers on calculating the tension in a rope twirling a 1.5 kg rock in a frictionless circle with a radius of 2.00 m, completing 100 revolutions in one minute. Key equations include Centripetal Force = Mass * (Speed^2) / Radius, which relates the tension in the rope to the centripetal force required for circular motion. The speed of the rock is derived from the circumference of the circle and the time taken for the revolutions. The tension in the rope is directly linked to the centripetal force needed to maintain the circular path of the rock.

PREREQUISITES
  • Understanding of Centripetal Force and its formula
  • Knowledge of circular motion concepts
  • Ability to calculate speed from distance and time
  • Familiarity with gravitational force calculations
NEXT STEPS
  • Calculate the speed of the rock using the formula Speed = Distance / Time
  • Apply the Centripetal Force formula to determine the tension in the rope
  • Explore the relationship between gravitational force and apparent weight in circular motion
  • Investigate the effects of varying mass and radius on tension in circular motion
USEFUL FOR

Physics students, educators, and anyone interested in understanding the dynamics of circular motion and tension in ropes during rotational activities.

psychfan29
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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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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
 
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...
 

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