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Rotating falling object

  1. Dec 6, 2013 #1
    1. The problem statement, all variables and given/known data

    A wet floor sign is pushed on the top and it starts falling. Just before hitting the floor the upper point of the sign sign hit a foot.
    The weight of the sign is 5 pounds (2.27 Kg or 22.24 N).
    The sign is 2 feet tall (0.61 m).

    2. Relevant equations

    With what force and energy would the foot be hit?

    3. The attempt at a solution

    Unfortunately there is no attempt at a solution, I don't have the knowledge yet. I tried to apply some equations I found but it didn't make sense to me.
  2. jcsd
  3. Dec 6, 2013 #2
    What were the equations you tried?
  4. Dec 7, 2013 #3


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    Are we to assume the sign has mass uniformly distributed along its height?
    Are we to assume the base of the sign does not slip? Or maybe that it's frictionless?
    Whatever the answers to those questions, asking for the 'force' of impact is meaningless. There will be a momentum of impact. The shorter the time the impact lasts, the greater the force, without limit. And one cannot know the duration of impact without details of the physical properties of the objects involved.
  5. Dec 7, 2013 #4
    I'm sorry I didn't give enough information, we should consider the mass uniformly and the base does not slip. About the force being meaningless I think it's the biggest error I made (try to find a force) but still I don't know how to calculate the velocity of the top I need for the momentum. Is it possible to calculate it with this informations or we need something more?
    Thank you!!!
  6. Dec 7, 2013 #5


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    Start with conservation of work. Can you write the equation for that?
    If the base does not slip then this becomes quite a complicated problem. You have to consider whether the object will become airborne. Is there some critical angle at which this will happen? Hints: consider centripetal acceleration, and which way the net force from the ground will point at the instant of losing contact.
  7. Dec 7, 2013 #6
    Fd=mgh=(mv^2)/2 is this right?
    What do you mean with "airborne", do you mean if we consider the air friction?? (I'm sorry, I'm an italian exchange student...)
    The centripetal acceleration would decrease from 9,81m/s^2 to 0, right? And at the same time the tangent acceleration would do the opposite process.
    Fd=(mv^2)/2 where F is the tangent force applied (gravity), d the distance travelled (centre of mass), m the mass and v the tangent velocity (centre of mass).
    I can use an average for F that would be my force of gravity at 45 degrees (since at 90 degrees the tangent acceleration from gravity is 0 and when it hit the ground/foot it would be the full acceleration of gravity) and find the distance (1/4 of circumference with radius 1 foot) and I have the mass.
    So I can get the velocity of the middle of the sign.
    Is it right?
    Then I could use the relation V=s/t and t=s/V to get the tangent velocity of the upper part:
    being the time equal t1=t2 so s1/V1=s2/V2
    I know the velocity and the space for the middle and I can calculate the space for the top.
    Finally when I have the tangent velocity of the top I can multiply it for the mass of the object and get the momentum.

    It it right???
    Thank you!!!
  8. Dec 7, 2013 #7


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    I mean that the object might not stay in contact with the ground. When it starts to fall from the vertical, it will build up a lot of horizontal and rotational speed but not much vertical speed. As it continues, more energy will go into increasing the vertical speed, but it already has a lot of rotational speed, so it might happen that the bottom of the object rises clear of the ground before the top hits the ground. There's no easy way to determine whether this could happen - you just have to go through the equations.
    No, centripetal acceleration is the acceleration that needs to result from the applied forces if an object is continue moving in a circle. As long as it stays in contact with the ground, the object is effectively rotating about that point. It therefore requires a centripetal force directed along its length towards that point. (It may also be accelerating tangentially, but that's another matter.)
    Do you know the equation for centripetal force or acceleration?
    If you think about the net force from the ground on the base of the object (the resultant of the normal and frictional forces), which direction will that be pointing in at the moment it loses contact from the ground (assuming that happens)?
    Not quite. For Fd to be the work done by F, d must be measured in the direction of F, i.e. vertically if F is mg. Also, the KE of the object is partly in the linear movement of its mass centre and partly in the rotation about that centre. Do you know the formula for rotational KE? The parallel axis theorem?
  9. Dec 8, 2013 #8
    The object doesn't go airborne and rotate around the bottom. :)

    I thought that there is just one force on the object (except the force that initiated the movement) during the process that's the acceleration of gravity. If the object is standing the acceleration will be discharged on the ground, while when it hit the ground the acceleration is accelerating it only tangentially since the sign is horizontal and the acceleration vertical. The centripetal force would be 0 only at the end, when the sign is hitting the ground.
    Centripetal force
    F=(mv^2)/2 where v is the tangental velocity.

    I know that d need to be in the same direction of F, it's why I wanted to use an average F:
    The tangental force of gravity is 0 in the beginning (all the force discharged on the ground) and equal to the force of gravity when it does hit the ground (the force of gravity is vertical and the object horizontal). So if we could use an average F that would be the one at 45 degrees. This would mean F/(square root of 2).

    I'm sorry, I don't know neither the rotational KE formula and the parallel axis theorem.

    Thank you!
  10. Dec 8, 2013 #9


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    If a uniform rod topples from a vertical position, having sufficiently frictional contact with the round that it does not slip, there is a point at which it will become airborne. (Specifically, if theta is the angle to the vertical, it will be when sin(θ) = 0.8.) From your description, this will apply to the OP.
    All the while it is still in contact with the ground, there is also a normal and frictional force from there.
    As long as one end is stationary and the other end is moving, there will be a centripetal force towards the stationary end. Remember that centripetal force is not an extra force - it is the resultant of other forces. That is, in order to keep rotating at radius r about the point of contact while moving at speed v, the sum of the gravitational force and the forces from the ground, when resolved parallel to the rod, must equal the centripetal force, mv2/r.
    Now, in most questions that deal with centripetal force it is being applied to a point mass. In the case of an object with some size, we need to bear in mind that not all parts of the object are the same distance from the axis of rotation. But for a uniform rod it turns out the same as if the mass were concentrated in the middle of the rod.
    No, that's the formula for KE. Centripetal force is mv2/r, where r is the radius of rotation (distance from axis). Equivalently, if the angular rate is ω then mrω2.
    Work/energy theorem: if all the forces that do work are conservative (and they are here) then KE+PE of the system is constant. If the rod has mass m and height h, how much PE does it lose in falling to the ground?
    Suppose, first, it were hinged to the ground at the base to prevent its becoming airborne, and that the top of the rod hits the ground at speed v: How fast (in terms of v and h) is it rotating when it hits the ground? Call the rotation rate ω.
    The moment of inertia of a uniform rod length h mass m about its centre is mh2/12. But this rod will rotate about its hinge on the ground, which is distance h/2 from the mass centre. The parallel axis theorem says we need to add m(h/2)2 to find the MoI about that point, giving I = mh2/3.
    The KE of a rotating object is Iω2/2. So in terms of v, m and h, how much KE does it have as it hits the ground?
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