The discussion revolves around solving complex pendulum equations, specifically focusing on energy conservation and tension in the string at various points of motion. Participants are examining the mechanics involved in a pendulum system, including centripetal force and gravitational effects.
There is ongoing exploration of the energy conservation equations, with some participants providing hints and guidance on how to approach the problems. Multiple interpretations of the setup and assumptions are being discussed, particularly regarding the reference points for height and the behavior of the pendulum at different positions.
Participants are working with images that are not clearly visible, which may affect their understanding of the problem. There is also a focus on ensuring that the reference points for height are consistent throughout the calculations.
BvU said:Number 3 first. Write down clearly your energy conservation equation.
thanks Sir. that's the most effective way for me to fulfill your order. btw, on number 3-3 I set KE for D negative, because it goes to the left. is that ok??BvU said:Unconventional presentation, but (quite !) good. So, on to number (5):
##mv^2\over r## of tension is needed to maintain the circular motion.
Once the component of ##mg## along the wire exceeds ##mv^2\over r## the circular motion is interrupted. This happens somewhere between B and D, at a height above C.
Introduce some coordinate, e.g. ##\theta## and work out ##v(\theta)## and from that ##T(\theta)##.
ok somehow maybe I understand. i'll try again thenBvU said:That's what I get for looking at the answers too summarily. Sorry.
No. Kinetic energy is never negative (*). The potential energy at D is not ##mg(2b-a)## but ##mg(a-2b)## because the height at D is lower than the height at A.
(*)
Both ##m## and ##v^2## are non-negative. For kinetic energy there is no direction, so 'because it goes to the left' does not apply.
ok somehow maybe I understand. i'll try again thenBvU said:That's what I get for looking at the answers too summarily. Sorry.
No. Kinetic energy is never negative (*). The potential energy at D is not ##mg(2b-a)## but ##mg(a-2b)## because the height at D is lower than the height at A.
(*)
Both ##m## and ##v^2## are non-negative. For kinetic energy there is no direction, so 'because it goes to the left' does not apply.
Draw a free body diagram with the mass somewhere between heights C and D, string still taut. Let the string make angle theta to the vertical here.Helly123 said:How do you know its above height C when mg exceeds mv^2/R ? And can you give me another hint to work on tetha to find v?
How if i used ΣF = ma on horizontal? Or i make all force in tetha direction?haruspex said:Draw a free body diagram with the mass somewhere between heights C and D, string still taut. Let the string make angle theta to the vertical here.
What is the speed of the mass, as a function of theta?
Write the ΣF=ma equation for the vertical direction.
Yes, sorry, that's better. ΣF=ma along the direction of the string.Helly123 said:Or i make all force in tetha direction?
But I am confused little bit. Let's say i make SigmaF = ma isharuspex said:Yes, sorry, that's better. ΣF=ma along the direction of the string.
You can choose to work in an inertial frame or in the frame of reference of the mass.Helly123 said:T + mg.cos tetha - centrifugal force = m.a
haruspex said:You can choose to work in an inertial frame or in the frame of reference of the mass.
In the mass' frame of reference, there is centrifugal force but the acceleration is, by definition, zero.
In the inertial frame, there is no centrifugal force but there is centripetal acceleration, i.e., the component of the resulting acceleration that is normal to the velocity.
The two equations that result are the same.
In the frame of reference of the mass, the mass is stationary.Helly123 said:Why a zero?
What happened to cos theta? You had that in post #42.Helly123 said:M.g - m.v^2/R = 0
Mgcos tetha - m.v^2/R = 0haruspex said:In the frame of reference of the mass, the mass is stationary.
What happened to cos theta? You had that in post #42.
For inertial force, there's centripetal force? Can you give me hint how to work using inertial force?haruspex said:You can choose to work in an inertial frame or in the frame of reference of the mass.
In the mass' frame of reference, there is centrifugal force but the acceleration is, by definition, zero.
In the inertial frame, there is no centrifugal force but there is centripetal acceleration, i.e., the component of the resulting acceleration that is normal to the velocity.
The two equations that result are the same.
No. When the angle is theta, how far is the mass above the peg? How far below point A does that make it?Helly123 said:V = sqrt g(a-b) cos tetha
Centripetal force is not an actual applied force, like gravity or tension. It is merely that component of the net force which is normal to the velocity. It produces the radial acceleration.Helly123 said:For inertial force, there's centripetal force?