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Sliding rod against a wall: with what velocity will its center of mass hit the ground?

  1. Nov 21, 2014 #1
    1. The problem statement, all variables and given/known data

    As a diagram any of the countless images on the internet serve well:

    ladder.gif

    Suppose the static friction was not sufficient to maintain the rod in a static equilibrium and it starts slipping/sliding. It has an initial angle theta with respect to the ground. Assume there is no dynamic friction and that the rod has uniform mass. The rod has a length L. With what velocity will it's center of mass hit the ground?

    2. Relevant equations
    The moment of inertia is useful in this case: [itex]I^*=\frac{1}{12}MR^2[/itex]

    [itex]\tau_{total} = I^* \alpha [/itex]

    3. The attempt at a solution

    I tried analyzing the forces and then looking at the torque as well.

    Y axis (positive up).

    [itex]N_{ground}-Mg=Ma_{CM_{y}} \Rightarrow N_{ground} = M(g+a_{CM_{y}})[/itex]

    X axis (positive right).

    [itex]N_{wall}=Ma_{CM_{x}}[/itex]

    Torques with respect to the center of mass (positive counterclockwise, or "out of the screen"):

    [itex]\tau_{total} = I^* \alpha = N_{ground}\frac{L}{2}cos\theta - N_{wall}\frac{L}{2}sin\theta = \frac{1}{12}ML^2\alpha[/itex]

    [itex] M(g+a_{CM_{y}})\frac{L}{2}cos\theta - Ma_{CM_{x}}\frac{L}{2}sin\theta = \frac{1}{12}ML^2\alpha[/itex] //substituted the normal forces

    [itex] (g+a_{CM_{y}})cos\theta - a_{CM_{x}}sin\theta = \frac{1}{3}\frac{L}{2}\alpha = \frac{1}{3} (a)_{t}[/itex] //a_{t} is the acceleration of a point at the end of the rod in the same direction as the same point's velocity

    I'm more or less stuck here. How can I relate the velocity of the center of mass with something I know?

    Thank you for your time.
     
    Last edited: Nov 21, 2014
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  3. Nov 21, 2014 #2

    haruspex

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    Consider the locations of the mass centre and the two endpoints as vectors. What equation relates them?
    You need to be careful not to make unwarranted assumptions in the equations. This is quite a tricky question. Bear in mind that the normal forces cannot go negative.
     
  4. Nov 21, 2014 #3
    I'm not sure what you mean. The vector v1 from the CM to the first endpoint, the one tangent to the wall, has a length L/2. The vector v2 from the CM to the point tangent to the ground is also L/2. If it were not a rod but a system with two particles each with some masses m1 and m2 but at the same position we could calculate the CM of the system without using integrals. It would be vCM (position)= (m1v1+m2v2)/(m1+m2) which is the midpoint if the masses are equal.

    Is this a warning for the future or are you telling me I have already made such an assumption? If so I do not see it.

    Thank you for your patience.
     
  5. Nov 21, 2014 #4

    haruspex

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    I was thinking in terms of an arbitrary origin. If ##\vec{x_a}##, ##\vec{x_{cm}}##, ##\vec{x_b}##, are the three vectors, what equation relates them? What do you get by differentiating that twice?
    A bit of both. Normal forces from one-sided contacts cannot go negative. This puts constraints on the applicability of your equations. You need to consider whether blindly following your equations to the point where the rod is flat on the ground will lead through a domain which violates this.
    To cover that, try to find out whether at some point a normal force can become zero.
     
  6. Nov 22, 2014 #5
    Oh, yes, now I understand what you meant. You can sum and rest Xcm to Xa and get the following equation, where x* is the position with respect to the cm.

    [itex]\vec{x_a}=\vec{x_a}+\vec{x_{cm}}-\vec{x_{cm}}=\vec{x_{cm}}+(\vec{x_{a}}-\vec{x_{cm}})=\vec{x_a}+\vec{x^*_a}[/itex]

    We can do the same thing with Xb and differentiating as you said you get similar identities for acceleration and velocity in vector form.

    In this problem ##\vec{a_{CM}}## is an unknown quantity and if we consider an endpoint, ##\vec{x_p}## we can apply the following after doing the adecuate derivation:

    [itex]\vec{a_p}=\vec{a_{cm}}+\vec{a^*_p}=\vec{a_{cm}}+\vec{a_t}[/itex]

    But these are vector quantities. I have the modulus of ##\vec{a_t}##, ##\frac{L}{2}\alpha##, in my equations, but I don't know how to describe the direction of ##\vec{a_t}##. It would be nice if I could use the velocities instead but I don't know ##\omega##.

    I can try looking at the formulas:

    [itex]N_{ground} = M(g+a_{CM_{y}})[/itex]

    g is a positive quantity given my sign convention, therefore the normal with the ground would become zero if a_CMy, which is negative given my sign convention again, would be greater than g in modulus. But I find this bizarre, how can an object fall faster than gravity pulls it if no other force is pushing it down?

    [itex]N_{wall}=Ma_{CM_{x}}[/itex]

    This would be negative only if a_CMx would be negative since mass is positive. That would mean the rod would be going upwards (left actually, but it can't penetrate the wall), tending to kiss the wall. I find this ridiculous, so in this context N_wall is always positive except when a_CMx is 0, which would correspond to the object at rest on the ground.
     
    Last edited: Nov 22, 2014
  7. Nov 22, 2014 #6

    TSny

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    A negative x-component of acceleration of the CM does not imply that the CM is moving toward the left.
     
  8. Nov 22, 2014 #7
    I think I see it now. It might just mean that the center of mass is decelerating if it was previously moving to the right.

    But if this is the case then it doesn't make sense to describe the normal with that equation, at least not in said case (decelerating horizontally). Does this mean my model, that is, the equations I considered, is inappropriate or incomplete? Is it at least useful or should I attack this problem from a different perspective entirely?

    If I am to use those equations I'll have to make the assumption that the acceleration if positive or at least delimit when it's applicable. But given how the problem is described it's basically assumed that the endpoints of the rod are always going to be in contact with the ground or the wall respectively, so there should be a normal force to take into account (even if its 0 in some cases, like when it hits the ground).
     
  9. Nov 22, 2014 #8

    ehild

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    The rod can lose contact with the wall. You can see if you do some little experiment with a pen and a book, for example. At the end, the pencil loses contact and slides on the ground.
    After detaching from the wall , Nwall =0, and aCMx =0. That means the velocity of the CM has a constant horizontal component.
    The rod is in contact both with the wall and the ground initially, and for some time after it. Solve the motion of the rod and see, at what position become one of the normal forces equal to zero.
     
    Last edited: Nov 22, 2014
  10. Nov 22, 2014 #9

    haruspex

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    Your equations are valid initially. From them you can calculate the angle at which the normal force from the wall becomes zero. You can use your equations to determine the complete circumstances at that point: the linear and angular moments and the position.
    From that point you can then apply different equations. As ehild mentions, the horizontal component of the velocity of the mass centre will now be constant, so you only have to consider the vertical velocity and the rotation.
    The next question is whether it ever loses contact with the ground. If it does the question becomes unanswerable without making assumptions about elasticity. If there's any elasticity, the mass centre only reaches the ground in the limit, and will have zero vertical velocity. At that point we could conclude that the question setter blundered.
     
  11. Nov 22, 2014 #10
    I don't know where to start when analyzing the general motion of the rod. For the moment I'll just try to set N_wall=0 and see what comes out.

    [itex]\tau_{total} = I^* \alpha = N_{ground}\frac{L}{2}cos\theta - N_{wall}\frac{L}{2}sin\theta = \frac{1}{12}ML^2\alpha[/itex]

    [itex](Ma+Mg)\frac{L}{2}cos\theta = \frac{1}{12}ML^2\alpha[/itex] //cancelled the second term since N_wall is null and substituted N_ground with its value; note that since a_CMx is null then a_CMy=a_CM=a

    [itex](a+g)cos\theta = \frac{1}{3}(\frac{L}{2}\alpha)=\frac{1}{3}a^*[/itex] // a is the acceleration of the CM, vertical only since a_CMx is zero

    Using the relation between CM and and object described with respect to the CM hinted at by another poster, we can get the following, where I have taken a_p to be the acceleration of the point in contact with the ground (since it's in contact with the ground it only has a horizontal component) which equals the acceleration of the CM plus the acceleration of the point with respect to the CM:

    [itex]\vec{a_p} = \vec{a} + \vec{a^*} \rightarrow a_{p}\vec{i} = a\vec{j}+a^*\vec{u}[/itex] //u unitary vector in the direction of a*, which is perpendicular to the rod
    [itex]a_{p}\vec{i} = a\vec{j}+\frac{L}{2}\alpha (\sin\theta \vec{i}+\cos\theta\vec{j})[/itex] this equality holds only if the components of the vectors are equal which gives the following system

    [itex]a_{p}=\frac{L}{2}\alpha\sin\theta[/itex]
    [itex]0=a+\frac{L}{2}\alpha\cos\theta[/itex]

    If I did all the previous steps correctly (which admittedly is a big if), then by dividing the two equations I get the following relation between the acceleration of the center of mass (vertical) and the acceleration of the point of contact (horizontal) but I don't feel it's very useful because I get the angle as a function of something I don't know:

    [itex]\tan\theta=-\frac{a_{p}}{a}[/itex]
     
  12. Nov 22, 2014 #11

    TSny

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    Does ##\vec{a^*}## have a centripetal component along the rod as well as a component perpendicular to the rod?

    Can you think of any quantity that is conserved during the motion?
     
  13. Nov 22, 2014 #12

    haruspex

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    Up until the point where it loses contact with the wall, geometry gives you the relationship between linear velocities and angular velocity. A conservation law relates those to the angle at that point. This will be much easier than solving differential equations.
    Having done that, you can derive the equation for the normal force from the wall at any given angle and deduce the angle at which that force becomes zero.
    Watch the signs. Which of the normal forces tend to increase the angle?
     
  14. Nov 22, 2014 #13
    Oh, I was wrong at least there then. I assumed that a^*=0.5L alpha (that's just the perpendicular component).
    Energy perhaps? It's 1 am here so I'll get on it tomorrow.
     
  15. Nov 23, 2014 #14

    ehild

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    Where is the centre of mass of a homogeneous rod? Is it not at the centre of the rod?
    Can you express the coordinates of the centre of the falling rod in terms of theta and the length of the rod?
    If you know the expression of the position of the CM in terms of theta, you get the components of velocity and the components of the acceleration by differentiating with respect to time.
     
  16. Nov 23, 2014 #15
    Using geometry as you said I get that the distance of the center of mass to the point O, which I take to be the corner formed between the wall and the ground, is constant, independent of the angle and even more, that distance is half the length of the rod, that is, L/2. The center of mass then, while the rod is touching the wall, follows a circular path around O. I will denote by lowercase L the distance from O to CM, that is, L/2=l.

    [itex]\vec{x_{cm}}=l\cos\theta\vec{i}+l\sin\theta\vec{j}[/itex] //we take O to be the origin
    [itex]\vec{v_{cm}}=-l\dot{\theta}\sin\theta\vec{i}+l\dot{\theta}\cos\theta\vec{j}[/itex]
    [itex]\vec{a_{cm}}=-l\dot{\theta}^{2}(\cos\theta\vec{i}+\sin\theta\vec{j})=-l\omega^{2}\vec{x_{cm}}=-\frac{l^2}{l}\omega^{2}\vec{x_{cm}}=-\frac{v_{lineal}^{2}}{l}\vec{x_{cm}}[/itex]
    [itex]v_{lineal}^2=v_{cm_y}^2+v_{cm_x}^2[/itex]

    Note that ##\theta(t)## doesn't necessarily grow linearly with time.

    Before going any further, these relations only help me while the rod is in contact with the wall. How do they help me find the final velocity of the center of mass, that is, when it hits the ground, in other words, when it's not in contact with the wall anymore?

    I don't think I understand the question. The greater the angle the greater sin is and the smaller cos gets. But it's difficult to analyze because the normals are themselves functions of the angle. If the position at rest of the rod is completely vertical then Nground is max and Nwall is null. If its completely flat on the ground, then Nwall is 0 because there are no other horizontal forces and it's at rest so Nground is max. Therefore Nwall in the case described should have a maximum value and Nground a minimal value, but I don't know if these observations help me.
     
  17. Nov 23, 2014 #16
    I come back to your question. Thinking it over again, I see that indeed it must have a centripetal component, if not the velocity with respect to the center of mass would not change direction. So ##\vec{a^*}=\vec{a_c}+\vec{a_t}##, where the modulus of a_t is L/2*alpha.

    [itex]\vec{a_p} = \vec{a} + \vec{a^*} \rightarrow a_{p}\vec{i} = a\vec{j}+\frac{v_{lineal}^2}{\frac{L}{2}}\vec{u_c}+a_t\vec{u_t}[/itex] //remember a is the acceleration of the center of mass, and that it is vertical

    [itex]a_{p}\vec{i} = a\vec{j}+2\frac{v_{lineal}^2}{L}(-\cos\theta \vec{i}+\sin\theta\vec{j})+\frac{L}{2}\alpha (\sin\theta \vec{i}+\cos\theta\vec{j})[/itex]

    Applying the same reasoning as before, the components on the left and on the right of the same vectors mush be equal and I get a system. But there is only one equation with many unknowns such as the acceleration of the point p (that is in contact with the ground), the angular acceleration alpha and the acceleration of the center of mass.
     
  18. Nov 23, 2014 #17

    TSny

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    OK. You can get a useful result from this by considering the ##\hat{j}## component. Llikewise, you can get another useful result by setting up a similar equation for the point of contact with the wall and considering the ##\hat{i}## component. Clearly, this result will only be valid as long as the rod is in contact with the wall. It might be a good idea to express ##v_{lineal}## in terms of the angular velocity ##\omega## of the rod.

    You can get these same results in a direct manner by following ehild's suggestion of writing ##x_{cm}## and ##y_{cm}## in terms of ##\theta## and taking time derivatives.
     
  19. Nov 23, 2014 #18

    TSny

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    Did you leave out some terms in the acceleration expression? Note that ##\dot{\theta}## is not a constant in the velocity expression.
     
  20. Nov 23, 2014 #19
    Ooops, that was a blunder.

    [itex]\vec{a_{cm}}=l\ddot{\theta}(-\sin\theta\vec{i}+\cos\theta\vec{j})-l\dot{\theta}^{2}(\cos\theta\vec{i}+\sin\theta\vec{j})[/itex]

    I have calculated it again taking into account what you said. The above is indeed correct. We can check by observing that the second parenthesis is the unit radial vector, and the modulus of that is the centripetal acceleration v^2/l pointing towards the origin. On the left we have a unit vector perpendicular to the "radial" vector with modulus R*alpha (R*angular acceleration with respect to the origin).

    Next I have an important question: when describing the acceleration of the center of mass I have used angular velocity and acceleration (theta dot and double dot). Theta is understood to be the angle of the position vector with the horizontal, because of the way I set things up. Originally theta was the angle of the rod with the ground but they are the same because of the geometry of the situation. Now then, when writing the moment of inertia I used a term alpha which is also angular acceleration, but with respect to the center of mass.

    Is ##\ddot{\theta}## and ##\alpha## the same in this situation?
     
  21. Nov 23, 2014 #20

    TSny

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    I think that's correct.

    They are the same up to a sign. When you set up your torque equation it looks like you are taking counterclockwise as positive for ##\alpha##. But, ##\theta## in the figure decreases for counterclockwise rotation of the rod. So, ##\alpha## is positive when ##\ddot{\theta}## is negative.
     
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