Uniform Ciruclar Motion (whirling cord)

  • Thread starter Thread starter KatieLynn
  • Start date Start date
  • Tags Tags
    Motion Uniform
AI Thread Summary
A 200g object is being whirled in a horizontal circle with a diameter of 2.4 meters at a rate of 3 revolutions per second. The velocity of the object was calculated as 22.3 m/s, and the acceleration was found to be 425.6 m/s². The confusion arose in calculating the tension in the cord, where the weight of the object was incorrectly included in the horizontal force calculations. The correct approach requires recognizing that the centripetal force acts horizontally, while the weight acts vertically. After correcting the mass to 0.2 kg, the tension in the cord was determined to be 85.1 N.
KatieLynn
Messages
64
Reaction score
0

Homework Statement



A 200g object is tied to the end of a cord whirled around on top of a horizontal and frictionless table top in a 2.4 diameter circle at 3 rev/s. Determine a) the velocity of the object b)t he acceleration of the object and c)t he tension in the cord.

Homework Equations



Vc = (2*pi*r)/T

Ac = (Vc^2)/R

Fnet = MA

Ft - Fg =MAc

The Attempt at a Solution



So for A) I got Vc = (2*pi*1.2)(1/3) which equals 22.3 m/s

Then for B) I got Ac=(22.6)^2/(1.2) which equals 425.6 m/s

Now here's where I get lost

I tried doing this but it doesn't come out right...

Fnet = MA

Ft - Fg =MAc

Ft - 196.2 = (20)(425.6)

which means Ft = 8708.2N, I know the answer should be Force of tension = 85.1N, I'm not sure what I'm doing wrong.
 
Physics news on Phys.org
200g=.2kg. Change to kg if you want the answer in Newtons. Watch your units.
 
KatieLynn said:

Homework Statement



A 200g object is tied to the end of a cord whirled around on top of a horizontal and frictionless table top in a 2.4 diameter circle at 3 rev/s. Determine a) the velocity of the object b)t he acceleration of the object and c)t he tension in the cord.

Homework Equations



Vc = (2*pi*r)/T

Ac = (Vc^2)/R

Fnet = MA

Ft - Fg =MAc

The Attempt at a Solution



So for A) I got Vc = (2*pi*1.2)(1/3) which equals 22.3 m/s

Then for B) I got Ac=(22.6)^2/(1.2) which equals 425.6 m/s

Now here's where I get lost

I tried doing this but it doesn't come out right...

Fnet = MA

Ft - Fg =MAc

Ft - 196.2 = (20)(425.6)

which means Ft = 8708.2N, I know the answer should be Force of tension = 85.1N, I'm not sure what I'm doing wrong.
you did fine in a and b. In c, however, you included the weight, but since the object is being twirled on a horizontal tabletop, the centripetal force and acceleration acts horizontally. The weight acts vertically.
 
PhanthomJay said:
you did fine in a and b. In c, however, you included the weight, but since the object is being twirled on a horizontal tabletop, the centripetal force and acceleration acts horizontally. The weight acts vertically.

Yes, there's that too. I missed it.
 
Ahh that you so much. I, for some reason, was converting 200g to 20kg, now that I switched it to .2kg and took out the force normal I got the right answer. Thank youuu :)
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Back
Top