• Support PF! Buy your school textbooks, materials and every day products Here!

Help with circular motion

  • #1
rock.freak667
Homework Helper
6,230
31

Homework Statement



A particle,A, of mass,m, hangs by a light inextensible string of length,a from a fixed point O. The string is initially vertical and the particle is then given a horizontal velocity,[itex]\sqrt{nga}[/itex]. Show that it will move round a complete vertical circle in a vertical plane provided [itex]n \geq 5[/itex]


Homework Equations



Centripetal force=[itex]\frac{mv^2}{r}[/itex]


The Attempt at a Solution



Well the resultant force of the tension in the string and the component of the weight provides the centripetal force.

[itex]F_c=T-W_{component}[/itex]

[tex]\frac{mv^2}{a}=T-mgcos\alpha....(*)[/tex]

If initially it is vertical then [itex]\alpha=0[/itex] (Doesn't really seem to help)

For the object to make a complete circle, then the string must be taut at the highest point (i.e. when [itex]\alpha=180,T\geq 0[/itex]

From (*)
[tex]T=\frac{mv^2}{a}+mgcos180 \Rightarrow T=\frac{m(\sqrt{nga})^2}{a}-mg[/tex]

So that

T=mgn-mg

For [itex]T \geq 0[/itex] then [itex]mng \geq mg \Rightarrow n \geq 1[/itex]

Which is not what I want to show.
 

Answers and Replies

  • #2
Doc Al
Mentor
44,882
1,129
Two points:
(1) At the top, the centripetal acceleration and the weight both point downward and thus have the same sign.
(2) [itex]\sqrt{nga}[/itex] is the initial speed at the bottom of the circle, not the speed at the top. How are those speeds related?
 
  • #3
rock.freak667
Homework Helper
6,230
31
Two points:
(1) At the top, the centripetal acceleration and the weight both point downward and thus have the same sign.
At the top the tension and weight point downwards and that provides the centripetal force..I hope


Doc Al;1713800(2) [itex said:
\sqrt{nga}[/itex] is the initial speed at the bottom of the circle, not the speed at the top. How are those speeds related?
oh..they aren't the same. I thought they were, I read given a horizontal velocity to mean that it moves with that velocity throughout the motion.

Well by the law of conservation of mechanical energy, (relative to the horizontal line through O)

Energy at the bottom=Energy at the top
[tex]\frac{1}{2}m(\sqrt{nga})^2-mga=\frac{1}{2}mV^2+mga[/tex]

and so

[tex]mV^2= mgan-4mga[/tex]

and putting that into the eq'n with tension

[tex]T=\frac{mgan-4mga}{a}-mg[/tex]
[tex]\Rightarrow T=mgn-5mg=mg(n-5)[/tex]
and [itex]m\neq 0,g\neq [/itex], the only way for [itex]T\geq0[/itex] is if [itex]n \geq 5[/itex]

Thanks for that!
 
  • #4
rock.freak667
Homework Helper
6,230
31
I forgot to put in the 2nd part of the question.

"If when the string OA reaches the horizontal, the particle A collides and coalesces with a second particle at rest also of mass m, find the least value of n for the vertical circle to be completed."

I think I must use the law of conservation of momentum here.

To get the velocity at A, use law of conservation of M.E.


[tex]\frac{1}{2}mnga-mga=\frac{1}{2}mV_{a}^2[/tex]

[tex]V_a=\sqrt{nga-2ga}[/tex]

Initial momentum = [itex]mV_a[/itex]
= [itex]m\sqrt{nga-2ga}[/itex]

Final momentum = [itex]2mV_f[/itex]

By the law of conservation of momentum.

[tex]2mV_f=m\sqrt{nga-2ga}[/tex]

[tex]V_f=\sqrt{nga-2ga}[/tex]

Correct so far?
 
  • #5
Doc Al
Mentor
44,882
1,129
[tex]2mV_f=m\sqrt{nga-2ga}[/tex]

[tex]V_f=\sqrt{nga-2ga}[/tex]
All good except for that last step--you left off the factor of 2 in the denominator.
 

Related Threads for: Help with circular motion

  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
10
Views
1K
  • Last Post
Replies
10
Views
1K
  • Last Post
Replies
3
Views
877
  • Last Post
Replies
1
Views
630
Replies
2
Views
2K
  • Last Post
Replies
9
Views
518
  • Last Post
Replies
4
Views
944
Top