
#1
Dec1905, 11:01 PM

P: 44

A particle of mass m slides down a frictionless surface through a height h and collides with a uniform vertical rod (of mass M and lenght d), sticking to it. The rod pivots about point O through an angle before momentarily stopping. Find the angle.
Now, I can find the tangential velocity of the particle when it hits the rod, and therefore, its omega. After this I just have absolutely no idea where to go with it. I'm assuming there is no friction between the rod and point O, and that the only force stopping this rod from continuing to spin around is gravity. I tried to say g = alpha * d but then thought gravity wasnt tangential at all points so it's not valid. It doesn't say whether the collision is elastic or inelastic, so I'm assuming inelastic. Momentum must be conserved, so at the collision, mrv = (Irod+Iparticle)(omega), but that won't tell me when it stops. Then I thought that perhaps the circumference through which the rodparticle system moves is h, so h=(theta)*d [arc length], but that's not right either. I have a feeling a torque may be involved here, but that could only give me an alpha not a theta. If anyone is willing to do it out it says the correct answer is arccos[ 1  (6m^2h)/(d(2m+M)(3m+M))], but I have no idea where that could even come from. I don't recognize any parts of it. 



#2
Dec2005, 12:43 AM

Sci Advisor
P: 905





#3
Dec2005, 12:49 PM

P: 44

That's exactly where I went wrong. What sort of potential energy do rotating things have? There's obviously a gravitational force acting on this, since it is vertically oriented. But how would that be worked into the equation. I mean, if a 1kg block were to fall down a slide of heigh 10m, it would have mgh = 1*9.8*10 = 9.8J of energy at impact with the rod. When would a pivoting rod of lenght d have a potential enery of 9.8J? Would it be when the rod is 10m off the ground again, but that's not rotational at all.
Well, if gravity has a force of mg on the rod, then it acts on the center of mass, so it's acting at a radius of d/2, since the rod is uniform, but also on the center of mass of the particle, which is at r. That creates a [tex]\tau = (\frac{d}{2}) \times (Mg) + d \times mg[/tex] and [tex]\tau = I \alpha[/tex] and [tex]\Delta\omega = \alpha t[/tex]. If I could find alpha, which I think I could from that torque, and I know that omega 2 = 0 and I found omega 1 from the momentum, then I could find out how long it takes to slow down, and then I could use [tex]\Delta \theta = \omega_{1}t + \frac{\alpha t^2}{2}[/tex], knowing omega 1, alpha, and t, and knowing theta 1 = 0, I could find theta 2, the answer. Hmmm....that, of course, supposes the torque is right. Then again, the conservation of energy would be easier (or atleast possible), but I'm missing something there. 



#4
Dec2005, 04:34 PM

Sci Advisor
P: 905

Particle slides into a rod and pivots
If the rod deflects so that the end raises a height h', then the centre of mass of the rod (if it is uniform) raises h'/2. So the potential energy of both together is mgh'+Mgh'/2.




#5
Dec2005, 06:16 PM

P: 44

Ok, I see why that would be true (I just made the angle 90 degrees and it made sense), so,
[tex]mgh = mgh' + \frac{Mgh'}{2} [/tex] because when the rod is momentarily at rest it lacks rotational kinetic energy, so its purely potential [tex]mh = mh' + \frac{Mh'}{2} [/tex] g is a common factor [tex]2mh = 2mh' + Mh' [/tex] multiply by 2 [tex]h'=\frac{2mh}{M+2m}[/tex] solve for h' Then I drew a picture, with the rod vertical and the length is d. At some angle [tex]\theta[/tex] the length is still d, but [tex]d'_{y}[/tex] is dh' so now I have a right triangle with adjacent side dh' and hypot of d, so [tex]cos\theta = \frac{dh'}{d}[/tex] plugging in h' from above, I wound up with [tex]cos\theta = \frac{d\frac{2mh}{M+2m}}{d} [/tex] [tex] cos\theta = 1  \frac{2mh}{d(M+2m)}[/tex] which has some parts of the answer in it, but it's missing the [tex]6mh^2 and (3m+M)[/tex] parts. 



#6
Dec2105, 04:03 PM

HW Helper
P: 508

1. Consurvation of angular momentum just before and after the collision.
[tex]m \sqrt{2gh} d = ( \frac {Md^2}{3} + md^2 ) \omega [/tex] and 2. conservation of energy just after the impect and at the maximum deflection point [tex] \frac{1}{2} (\frac {Md^2}{3} + md^2 ) \omega^2 = (m + \frac {M}{2} ) gd (1 cos \theta ) [/tex] eliminate w MP 


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