Finding tension in cord when dart-block pendulum swings back down

[vincentchan]

do you know the formulas of potential energy and kinetic energy, if no, here is the formulas:
PE= mgh (h is the height between the object and the floor)
KE= 1/2 mv^2
and the total energy is PE+KE
since energy cannot come out from nothing, the total energy for the system at anytime is the same... just compute the total energy at the lowest point and the highest point... set them equal... vala, you have the formulas of conservation of energy...

 

[dextercioby]

Simple.In this case,the law of total energy conservation can be applied and it reads,for two arbitrary states.

$$KE(1)+PE(1)=KE(2)+PE(2)$$
The explicit expression i posted can be deduced from the geometry of the figure.
You chose the lowest point of the circular trajectory as to have 0 potential energy and the upmost to have only potential energy and the kinetic one to be zero.Analyzing the geometry,u can reach the formula almost immediately.

Daniel.

 

[physicsgirl101]

Thanks, I'll let you know if i get the right answer when I do it out----THANKS!

 

[physicsgirl101]

Okay,the two bodies which are joint together are in a circular movement round the point of suspension "O".
The Second law of Dynamics for this body,projected onto the radial direction reads
$$T-(m+M)g\cos\theta =\frac{(m+M)v^{2}}{l}$$ (1)
,where $\theta$ is the angle at center,"T" is the tension in the in wire and [itex]\frac{v^{2}}{l} [/tex] is the radial acceleration.<br /> <br /> Now,write down the law of energy conservation between the highest point (in which the system has only potential energy) and the lowest point in which the system has only kinetic energy:<br /> $$\frac{(m+M)v^{2}}{2}=(m+M)gl(1-\cos\theta_{max})$$ (2)<br /> Note that this formula yields the maximum linear velocity the body (total) can have.Therefore,combining (1) and (2),taking into account that in both formulas,the subscript 'max' for angles and velocity interviens<br /> $$T_{max}=(m+M)g+2(m+M)g(1-\cos\theta_{max})$$(3)<br /> <br /> ,which can be simplified to<br /> $$T_{max}=(m+M)g(3-2\cos\theta_{max})$$(4)<br /> <br /> The reason for why I've set $\theta=0$ is that u must maximize the function T.Which means maximize the velocity (which is done in formula (2)) and maximize the gravity as well,which is done by setting $\cos\theta =1$ or,equivalently,$\theta=0$<br /> <br /> Daniel.[/itex]

$<br /> For your equation for the second law of dynamics... is it basically saying the sum of the forces equals mass x (radial)acceleration which is also equal, in this case, to Tension minus Fg (minus because Fg is negative). .. ? So you set them equal... (?)$

 

[physicsgirl101]

YAY! I did it on my own using what you had said about setting energies in different positions equal and I got the same answer as yoU! thanks.

 

[dextercioby]

The second law reads in vector form
$$m\vec{a}=\vec{G}+\vec{N}$$

Project this law on radial direction.U'll find my equation.


Daniel.

 

[physicsgirl101]

The second law reads in vector form
$$m\vec{a}=\vec{G}+\vec{N}$$

Project this law on radial direction.U'll find my equation.


Daniel.

is one of the equations set up T-(M+m)g = ...(that's G + N, right?)
because (M+m)g is negative... ?