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Lever system with a moving mass

  1. Sep 23, 2008 #1
    Hi all, I thought of an interesting problem that I don't know how to approach (it's been 8 years since my last physics class in undergrad studies).

    Here is a crude depiction of the problem:

    _____■_____ _____●_____ lever
    .................↑................................... ground

    Assume the cube and sphere (obviously depicted here as a square and circle, respectively) have equal mass and are equidistant from the fulcrum, so that the lever is in a state of equilibrium.

    Now suppose that some additional external stimulus acts upon the sphere to disrupt the equilibrium such the sphere begins to roll down the lever.

    How would I calculate how long it would take the the circle to reach the ground and its terminal velocity when it hits the ground?

    There must be some differential equation(s) that describes the system.

    Thanks for any insight you can offer.
  2. jcsd
  3. Sep 23, 2008 #2
    OK here's a first attempt.

    The first thing to thing about is how a sphere accelerates on a static slope inclined at angle [tex] \beta [/tex] to the horizontal.

    If you draw a diagram of this, you can resolve the weight of the sphere, Mg, into a component parallel to the slope [tex] Mgsin(\beta) [/tex] and a component normal to the slope [tex] Mgcos(\beta) [/tex].

    I assume rolling without slipping throughout all of this btw. This leads us to two equations:[tex] Mgsin(\beta) -f=Ma_c [/tex], where f is the frictional force, and [tex] a_c [/tex] is the acceleration of the centre of mass. This is just straightforward Newton II applied to centre of mass of sphere. The second equation comes through considering the torque on the sphere due to the friction, [tex] fR=I_c \alpha= \frac{2}{5}MR^2 \alpha [/tex]. Where I have used the moment of inertia of a sphere about its centre of mass, [tex] I_c=\frac{2}{5}MR^2 [/tex] Now if rolling without slipping is occuring [tex] \frac{a_c}{r}= \alpha [/tex] , subbing this into the previous equation we finally obtain: [tex] fR= \frac{2}{5}MR a_c [/tex].

    Going back to the first equation: :[tex] Mgsin(\beta) -f=Ma_c [/tex],implies :[tex] f=M(gsin(\beta) -a_c) [/tex]. Equating the two equations:

    [tex] M(gsin(\beta) -a_c)R=\frac{2}{5}MR a_c [/tex] which leads us to [tex] a_c=\frac{5}{7}gsin(\beta) [/tex].

    So thats for a sphere sliding down a static slope of angle beta!

    I assume in your question the cube is nailed down or something, otherwise it literally will be a nightmere. The only way I can think of proceeding, is now to try and figure out how beta changes, so consider the torque and moment of intertia of the whole system.

    Take this next bit with a pinch of salt as Im not sure if Im barking up the wrong tree here:

    If the mass of cube is [tex]m _1 [/tex] and the systems are in equilibrium when [tex] m_1 l_1=M l_2 [/tex] where M is mass of the sphere.Net torque is
    [tex] \tau=m_1 l_1-M(l_2+x) [/tex] where x is the distance the sphere has slid toward end of plank from its equilibrium distance [tex] l_2 [/tex].

    Then since equilibrium implies [tex] m_1 l_1=M l_2 [/tex] . [tex] \tau=-Mx[/tex]. Now [tex] \tau=I (x) \ddot{\beta} [/tex]. But now the moment of inertia of the system is also dependent on x: [tex] I(x)=m_1 (l_1)^2+M(l_2+x)^2 [/tex]. Anyway you get an equation for beta:

    [tex] \ddot{\beta}=-\frac{Mx}{m_1 (l_1)^2+M(l_2+x)^2 } [/tex]. If you can somehow solve that to get the angle as a function of x, you can plug beta back into the static slope solution [tex] a_c=\frac{5}{7}gsin(\beta) [/tex] to get the acceleration as a function of x.

    Like I say not sure if the last bit is wholly correct, just a first stab at it. Now I've done all that Im thinking Lagrangian might be the way forward
  4. Sep 23, 2008 #3
    yeah definitley get the Lagrangian out.

    [tex] T=\frac{1}{2} M \dot{x}^2+\frac{1}{2} I \frac{\dot{x}^2}{R^2}+\frac{1}{2}[m_1 (l_1)^2+M(l_2+x)^2] \dot{\beta}^2 [/tex]

    [tex] U=m_1 g l_1 sin(\beta)-Mg(l_2+x)sin(\beta) [/tex]

    So that the Lagrangian [tex] L=\frac{1}{2} M \dot{x}^2+\frac{1}{2} I \frac{\dot{x}^2}{R^2}+\frac{1}{2}[m_1 (l_1)^2+M(l_2+x)^2] \dot{\beta}^2-m_1 g l_1 sin(\beta)+Mg(l_2+x)sin(\beta) [/tex]

    Applying the Euler Lagrange equations you end up with two coupled differential equations:

    [tex] \frac{7}{5}M\ddot{x}=M(l_2+x)\dot{\beta}^2+Mgsin(\beta) [/tex] which as you can see reduces to the equation I derived in my first post [tex]\frac{7}{5}M\ddot{x}=Mgsin(\beta) [/tex] when beta is a constant.

    and the second equation:

    [tex] [m_1 (l_1)^2+M(l_2+x)^2] \ddot{\beta}+2M(l_2+x)\dot{x}\dot{\beta}=gcos(\beta)[M(l_2+x)-m_1 l_1] [/tex]
    Last edited: Sep 23, 2008
  5. Sep 25, 2008 #4
    What a fantastic solution we have here. But I'm already stuck with the first equation of motion. Sorry that I'm not familier with the physics of rolling.
    My question is why does the friction act up the slope? The sphere is rolling without slipping. The velocity of the point on the sphere which is in contact with the slope is up the slope. The friction opposes the motion. So it must act downward at the point of contact with the sphere. :confused:
  6. Sep 26, 2008 #5
    The gravitational force acts on the sphere at its centre of mass (which is in this case its actual centre), take a coin and incline one of your fingers to a flat surface, now drag the penny at its centre down your finger. Try preventing it rolling while sliding it down, you should be able to see from this that friction is acting up the slope and trying induce rolling of the penny in the corresponding direction.

    This is contrary to say if you held an object spinning like a buzz saw, near a plane surface, this object would already have angular velocity and friction would obviously oppose this. Hence if you take a buzz saw like object, spinning in a direction corresponding to the penny rolling down a slope's ang velocity, then bring it near to the slope, friction would indeed act on this down the slope.

    Hope that makes sense.
  7. Sep 26, 2008 #6
    Infact another way to think about it, is for rolling without slipping the velocity of the centre of mass must always be [tex]R \omega [/tex], so at any instant the velocity of the point in contact with the slope at the perimeter of the penny, is moving uphill with [tex]v=R \omega [/tex] and the centre of mass of the penny is moving downhill also with [tex] v=R \omega [/tex]. Thus for the instant the perimeter point is in contact with the surface it actually zero velocity with the slope.

    So then you may ask well if the relative velocity is zero how is there any friction at all? (ignoring rolling friction which is something different entirely due to deformations).Then you have to remember that the centre of mass of the penny is accelerating, so a moment later the coin's centre of mass is going a tiny bit faster than [tex] R \omega [/tex], which is almost like a infinitesimal slipping, until the uphill friction induced by this corrects, and brings the perimeter upto the new velocity.

    Something even more confusing is if you role a coin uphill (without slipping) and gravity pulling downhill, the friction is AGAIN UPHILL. This is because the centre of mass is deccelerating, which means the perimeter velocity (bit in contact with surface) will start to exceed the centre of mass velocity. Then its like the buzz saw! instead of the bit in contact having zero net velocity w.r.t the slope, it actually has a non zero velocity in a downhill direction, and just like the buzz saw friction will act up hill against this. If this didnt happen the penny would eventually stop near the top of the hill, but still be spinning!

    If you pull a coin up with a rope through its middle this is not the case, then friction is downhill.

    Everything basically comes down to how fast the edges are moving opposite to the centre of mass's motion, and thus how fast their net velocity is w.r.t the slope.

    This last result took me a while to get my head round, I encourage you to find a coin.
  8. Sep 26, 2008 #7
    Thanks h0dgey84bc for the explaination. I like your second explanation. Just valuing the saying of Henri Poincare. "It is with logic that one proves; it is with intuition that one invents''

    Now I see. The net velocity of a rolling point is the sum of rotating velocity and the velocity of the centre of mass.

    To make use of the Newton second law we need to know all the forces acting on the body. Can I just generalized that the direction of the friction is just opposite to that of the net forces acting on the body.
    e.g. if the sphere is rolling down/up the slope then the component of the gravitational force parallel to slope is acting downward. So the friction is up the slope.
    e.g If we pull a coin up with a rope, the net force is upward. So the friction is down the slope.

    One trivial question: Can we roll a sphere on a frictionless surface?
  9. Sep 27, 2008 #8
    yeah pretty much, and if the ball is free rolling up the hill the gravitational force is still downwards, friction is upwards still.

    No rolling, the sphere would just slide down.
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