Pendulum/Circular Motion Problem

  • Thread starter Jay0078
  • Start date
  • Tags
    Motion
In summary: I think it's called a "radial distance." So r would be the distance that the mass has traveled.No, that's the speed needed to reach that height. But if the string is to stay taut it must be supplying (some nonzero portion of) the centripetal force. Consider the forces when in this position. What speed is it moving at if it's still moving in a circle but the string tension is just reaching zero.If you are releasing the mass from height L, it will have a speed of v_0 at the radial distance r.
  • #1
Jay0078
11
0

Homework Statement



A small ball of mass m is attached to a very light string of length L that is tied to a peg at
point P. What is the magnitude of the horizontal velocity that must be applied to the ball
so that it swings up and lands on the peg? Your answer can only contain the given
information and any appropriate physical constants (such as g).
If you have the acceleration of gravity g in your answer do not
substitute numbers for it. You may introduce other variables to
help you reach a solution but your final solution can only
contain the given variables and g.
(SEE ATTACHED)

Homework Equations



Conversation of Mechanical Energy?

The Attempt at a Solution



My initial thought was to set ΔK=ΔP since gravity is a conservative force and the only displacement is vertically, the length of the string. That gave me an initial velocity of √2gL
 

Attachments

  • Screen Shot 2014-04-12 at 12.36.03 PM.png
    Screen Shot 2014-04-12 at 12.36.03 PM.png
    7.1 KB · Views: 544
Physics news on Phys.org
  • #2
Your solution only gets the ball horizontal with the peg. You need to give it more speed to cause it to land on top of the peg.
 
  • #3
Exactly. Thanks for the reply. I found an initial velocity to make a complete revolution also, which is not what I'm looking for. Any ideas anyone?
 
  • #4
And what velocity is that?
 
  • #5
Jay0078 said:
Exactly. Thanks for the reply. I found an initial velocity to make a complete revolution also, which is not what I'm looking for. Any ideas anyone?

Clearly the answer lies between those two extremes.
You need the string to go slack at some point above the horizontal, but before the vertical.
Suppose it goes slack at angle θ above the horizontal. What initial velocity would lead to that (as a function of θ)? What will the subsequent trajectory be?
 
  • #6
paisiello2 said:
And what velocity is that?

√(2gL) ≤v_0≤ 2√(gL)
 
  • #7
haruspex said:
Clearly the answer lies between those two extremes.
You need the string to go slack at some point above the horizontal, but before the vertical.
Suppose it goes slack at angle θ above the horizontal. What initial velocity would lead to that (as a function of θ)? What will the subsequent trajectory be?

Could I relate that to the tension in the string? Or centripetal acceleration?
 
  • #8
Jay0078 said:
√(2gL) ≤v_0≤ 2√(gL)
How did you come up with that?
 
  • #9
Jay0078 said:
Could I relate that to the tension in the string? Or centripetal acceleration?

Yes, when the string goes slack the tension becomes________.
 
  • #10
dauto said:
Yes, when the string goes slack the tension becomes________.


Right, I'm just trying to find a way to relate them mathematically without introducing unnecessary variables because the solution can only contain the given variables.
 
  • #11
Jay0078 said:
Right, I'm just trying to find a way to relate them mathematically without introducing unnecessary variables because the solution can only contain the given variables.
It is often appropriate to introduce extra variables, then find enough equations to eliminate them. I would certainly recommend introducing one for the angle at which the string goes slack.
 
  • #12
I'm needing some direction. I realize that at some point after the horizontal the string must go slack; at this point it seems to be a projectile motion problem with a displacement in the x direction (L). Any thoughts?
 
  • #13
haruspex said:
It is often appropriate to introduce extra variables, then find enough equations to eliminate them. I would certainly recommend introducing one for the angle at which the string goes slack.

How would I do that?
 
  • #14
haruspex said:
It is often appropriate to introduce extra variables, then find enough equations to eliminate them. I would certainly recommend introducing one for the angle at which the string goes slack.

Vo=√(2gL(1-cosΘ))

How am I coming along? I'm working on finding a way to relate Θ back to the velocity needed to get the ball back to the peg.
 
  • #15
Jay0078 said:
Vo=√(2gL(1-cosΘ))
No, that's the speed needed to reach that height. But if the string is to stay taut it must be supplying (some nonzero portion of) the centripetal force. Consider the forces when in this position. What speed is it moving at if it's still moving in a circle but the string tension is just reaching zero.
 
  • #16
I got a ratio of L to r, but I would not swear it is correct. Is the answer given in the back of the book or something, to work towards? (actually looking at your thumbnail, I find I did a related (equivalent) problem, not this one)
 
  • #17
mpresic said:
I got a ratio of L to r, but I would not swear it is correct. Is the answer given in the back of the book or something, to work towards? (actually looking at your thumbnail, I find I did a related (equivalent) problem, not this one)
r? What distance is that?
 
  • #18
I did the problem where it is released horizontally from height L, and hits a peg at L - r, so that r is the radius of the path. The mass does not make the orbit with radius r but the tension goes to zero at some angle. After the tension hits zero, the mass goes on a parabolic path to the peg. This may make the problem even harder.
 
  • #19
haruspex said:
No, that's the speed needed to reach that height. But if the string is to stay taut it must be supplying (some nonzero portion of) the centripetal force. Consider the forces when in this position. What speed is it moving at if it's still moving in a circle but the string tension is just reaching zero.

I got v=√(gL) for the tension just going to zero at the top.
 
  • #20
mpresic said:
I got a ratio of L to r, but I would not swear it is correct. Is the answer given in the back of the book or something, to work towards? (actually looking at your thumbnail, I find I did a related (equivalent) problem, not this one)

No, I have searched and searched. However, I did find a tarzan-physics problem that had to do with rotation and projectile motion. It seemed fairly complex though it may be helpful.

http://arxiv.org/pdf/1208.4355.pdf
 

Attachments

  • Screen Shot 2014-04-13 at 10.01.12 PM.png
    Screen Shot 2014-04-13 at 10.01.12 PM.png
    18.3 KB · Views: 479
  • #21
Jay0078 said:
I got v=√(gL) for the tension just going to zero at the top.
Sure, but we don't need it taut at the top.
In your post #14, you seem to have taken theta as the angle traversed from starting position, so I'll stick with that.
Draw a free body diagram for the mass when the string tension just reaches zero, so it's still moving in a circle (just). What forces act on the mass? Suppose it is moving at speed vθ. What is the component of its acceleration towards the anchor point? What equation does that give you?
 

1. What is a pendulum?

A pendulum is a weight suspended from a fixed point so that it can swing freely back and forth under the influence of gravity. It is commonly used to keep time in clocks and has been studied by scientists for centuries to understand the principles of motion.

2. How does a pendulum work?

A pendulum works by converting the potential energy of its raised position into kinetic energy as it swings back and forth. The force of gravity pulls the pendulum downward, causing it to accelerate and gain speed as it moves towards the bottom of its swing. As it reaches the bottom, the kinetic energy is at its maximum and the pendulum begins to slow down due to the force of gravity pulling it back towards its resting position. This back and forth motion continues until the energy is eventually dissipated due to friction and air resistance.

3. What factors affect the period of a pendulum?

The period of a pendulum, or the time it takes to complete one full swing, is affected by the length of the pendulum, the mass of the weight, and the strength of gravity. The longer the pendulum, the longer the period. Similarly, a heavier weight will have a longer period than a lighter weight. The strength of gravity also plays a role, as a higher gravitational force will result in a shorter period.

4. Is the motion of a pendulum considered circular motion?

Yes, the motion of a pendulum is considered circular motion because the weight follows a circular path as it swings back and forth. This path can be seen by tracing the motion of the weight over time, and it can also be described using circular motion equations and principles.

5. How is circular motion different from pendulum motion?

Circular motion refers to the motion of an object following a circular path at a constant speed, while pendulum motion involves a back and forth swinging motion with a changing speed. Additionally, circular motion does not involve the influence of gravity, while pendulum motion relies on the force of gravity to keep the pendulum swinging.

Similar threads

  • Introductory Physics Homework Help
Replies
9
Views
636
  • Introductory Physics Homework Help
Replies
19
Views
950
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
14
Views
436
  • Introductory Physics Homework Help
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
626
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
4K
  • Introductory Physics Homework Help
Replies
9
Views
2K
  • Introductory Physics Homework Help
Replies
14
Views
1K
Back
Top