Circular motion grade 12 physics

  • Thread starter Specter
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  • #1
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Homework Statement


Keys with a combined mass of 0.100 kg are attached to a 0.25 m long string and swung in a circle in the verical plane.

a) What is the slowest speed that the keys can swing and still maintain a circular path?

b) What is the tension in the string at the bottom of the circle?

Homework Equations


Fc=mv2 /r

The Attempt at a Solution


I haven't learned about conservation of energy yet but it looks like I will need it to solve this problem. I'm stuck on part b but I will post my answer to part a and b.

A) When they keys are at the top of their path the only force that is acting on them is gravity.

Fc=Fg=mg
Fc=mv2 /r
g=v2 /r
v2 =√(9.8m/s)(0.25 m)
v= 1.565 m/s

Therefore the minimum velocity that is required to keep the keys swinging in a circular path is 1.565 m/s.

Part b.

I have only learned about uniform circular motion. If gravity accelerates the keys on the way down and slows them on the way up, wouldn't that mean this question is not uniform circular motion? I tried my best to solve it but I haven't actually learned any of this yet so it will probably have mistakes.

My first attempt is solving it as uniform circular motion because thats what I learned in the lesson:

Fc= mv2 /r
Ft-Fg = mv2 /r
Ft-mg= mv2 /r
Ft= mv2 /r +mg
Ft= (0.100kg)(1.6m/s)2 /0.25m +(0.100kg)(9.8m/s2)
Ft=2.0 N

I don't think this is right because the velocity at the top and bottom would be different but the only way to find the velocity at the bottom would be to use conservation of energy which I haven't learned yet...

Attempt 2 assuming it isn't uniform circular motion:

The top of the circle will be 0.50m and the bottom will be 0m.

Top of the circle

Eg=mgh
=(0.100)(9.82)(0.50)
=0.49 J

Bottom of the circle

Eg=mgh
=(0.100)(9.82)(0)
=0 J

0.49 is the GPE at the top of the circle.

I can calculate the kinetic energy to find the potential energy at the bottom

Ek=1/2 mv2
=1/2 (0.100)(1.565)2
=0.12 J

Etotal = Eg+Ek
=0.49 J + 0.12 J
= 0.61 J

The bottom of the circle is set to 0 therefore the GPE at the bottom will equal 0 and the kinetic energy at the bottom will equal 0.61 J.

Use kinetic energy to find velocity

Ek= 1/2 mv2
v2=√2(0.61)/0.100
v=3.49 m/s

Now using velocity to find the tension in the string

Fc=Ft-Fg
Ft-Fg=mv2 /r
Ft=mv2 /r +mg
Ft=(0.100)(3.49)2 /0.25 +(0.100)(9.82)

Therefore the tension in the string at the bottom of the circle is 5.85 N.
 

Answers and Replies

  • #3
120
8
Are you allowed to iuse calculus to solve part b?
Nope. I am taking a calculus course also, but I think they want me to use what I learned in the lesson and nothing else. How would I use calculus to solve this problem?
 
  • #4
jbriggs444
Science Advisor
Homework Helper
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the only way to find the velocity at the bottom would be to use conservation of energy which I haven't learned yet...
Conservation of energy is by far the easiest way to determine the velocity at the bottom of the path. Do you know about gravitational potential energy?
 
  • #5
120
8
Conservation of energy is by far the easiest way to determine the velocity at the bottom of the path. Do you know about gravitational potential energy?
I very briefly read about gravitational potential energy but I haven't learned about it in the lesson where this question came from. I used GPE=mgh to find the gravitational energy at the top and the bottom of the circle in my second attempt.
 

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