Solving Pendulum Pulley System w/ Gravity: H.G.'s Question

[hgetnet]

This is not a homework problem. It was a problem posed by my professor at the end of the semester but had me puzzled for a while. I was able to satisfactorily solve it for the case when gravity is not present but I was not so successful with gravity.

The question is: to determine the trajectory of the mass m1, in terms of
-mass (m2),
-initial length (initial radius r0) of the pendulum and
-the initial angle (theta0) that the string makes with respect to the side of the table and
-the initial distance of m2 from the pulley nearest to it (x0).

Solution would be appreciated. Or, explanation of why a college freshman can not solve such a problem would also do.

Thank you.
H.G.

 

[tiny-tim]

Welcome to PF!

Hi H.G.! Welcome to PF!

Just use conservation of energy

(and remember the speed of m2 is the same as the rate of increase in length of the hanging part of the string)

 

[hgetnet]

Thanks Tiny Tim.

--- I had this question posed to me few years back. Sorry if I am not clear on the problems I ran into.

(BTW: Pendulum - m1 - is not attached to the pulley but is descending to the ground as it pulls m2 with it while at the same time swinging)

 

[tiny-tim]

The thing that I could not figure out is how to express the velocity of the pendulum as it descends and swings, therefore its height and its relationship to the velocity of m2.

Just call the angle θ, and express the velocity in terms of r' and rθ'

 

[hgetnet]

Just call the angle θ, and express the velocity in terms of r' and rθ'

so the velocity of m2 would be dr/dt.

what is the velocity of m1? it can not be dr/dt because it is also swinging. What value of velocity would u use for the calculation of its (m1's) kinetic energy?

P.S> tiny_tim: *I appreciate the quick response*
p.s.s> I will post my solution for the case when there is no gravity

 

[tiny-tim]

what is the velocity of m1? it can not be dr/dt because it is also swinging. What value of velocity would u use for the calculation of its (m1's) kinetic energy?

You must learn this … radial speed = dr/dt, tangential speed = r dθ/dt.

 

[hgetnet]

You must learn this … radial speed = dr/dt, tangential speed = r dθ/dt.

so, we have non-linear differential equation in two variables (not counting t)..

$$\Delta$$U + $$\Delta$$KE = 0;

m1 gr( cos ($$\theta$$) - cos ($$\theta0$$) ) + 0 + 0.5*(m1[(dr/dt)$$^{2}$$ + (rd$$\theta$$/dt)$$^{2}$$ ] + m2* (dr/dt)$$^{2}$$) = 0;

0 = k1 + k2 * cos($$\theta$$) + k3 * r *$$^{'}$$$$^{2}$$$$\theta$$ + k4 *(r$$^{'}$$)$$^{2}$$ ;

What other relationship does exist that would turn this into a simultaneous equation?

 

[tiny-tim]

Hi hgetnet!

so, we have non-linear differential equation in two variables (not counting t)..

$$\Delta$$U + $$\Delta$$KE = 0;

m1 gr( cos ($$\theta$$) - cos ($$\theta0$$) ) + 0 + 0.5*(m1[(dr/dt)$$^{2}$$ + (rd$$\theta$$/dt)$$^{2}$$ ] + m2* (dr/dt)$$^{2}$$) = 0;

0 = k1 + k2 * cos($$\theta$$) + k3 * r *$$^{'}$$$$^{2}$$$$\theta$$ + k4 *(r$$^{'}$$)$$^{2}$$ ;

What other relationship does exist that would turn this into a simultaneous equation?

Yes, we need one more equation …

which we can get by taking components of force and acceleration for the swinging mass either in the radial or the tangential direction …

though I must admit, having tried it, that I can't actually see any simple solution to the combined equations.

hmm … perhaps that's why your professor posed it at the end of the semester? ​