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Lost on rotational motion problems

  1. Apr 1, 2005 #1
    We have recently started studying rotational and rolling motion, Ive struggled with this more than anything, Im pretty lost on these problems:

    1) Consider a thin rod of mass 3.4 kg, length 6.8m, and uniform density. The rod is pivoted at one end on a frictionless horizontal pin. The rod is initally held in horizontal position but eventually allowed to swing down. What is the angular acceleration at the instant it makes a 48 degree angle with the horizontal?

    So far Ive done this.

    Line density of rod = 3.4kg/6.8m =0.5 kg / m

    Moment of inertia = Integrate over 0 to 6.8 ((X^2) * 0.5 dx) = (0.5/3) * 6.8^2 =52.4 kg m^2

    Then using torque:
    (3.4) * (9.8) * (0.5 * 6.8) * sin(48) = 52.4a (where 'a' is angular acceleration)

    This comes out to 1.6066 m/s which converted to radians /s = 1.6066 m/s / 6.8 = .23626 rad/s^2

    but this is wrong, what is going wrong in this calculation?




    A 0.13kg basketball has 0.145m diameter and may be approximated as a thin sphereical shell. The moment of inertia is I = (2/3)m(R^2) and coefficient of friction 0.33.
    Starting from rest how long will it take a basketball to roll, without slipping, 6.18m down an incline that makes and angle of 30 degrees with the horizontal?

    Ive tried many different ways of going about this problem but Im still lost. Ive been trying to somehow solve with:
    KE=(1/2)mv^2 + (1/2) I*w^2

    I= (lambda)*m*r^2

    Im having no luck with these equations, and Im having a hard time just grasping the concepts of the problem, especially how to take friction into account, can anyone offer any insight?


    Thanks for any help.
     
  2. jcsd
  3. Apr 1, 2005 #2

    Doc Al

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    Prob 1:
    Good.

    The angle is 48 degrees below the horizontal; you're mixing up sine and cosine.

    Prob 2:

    This is good. But how do v and [itex]\omega[/itex] relate to each other if the ball rolls without slipping?

    Now apply conservation of energy: the loss in PE equals the gain in KE.

    You're not doing so bad! All the friction does is make the ball roll. Since there is no slipping, no energy is lost to friction.
     
  4. Apr 1, 2005 #3

    Doc Al

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    More on problem 2:

    Using energy methods is just one way of solving this problem! (The idea would be to find the speed at the bottom, then use kinematics to calculate the time of travel.)

    But you can also use Newton's laws to find the acceleration down the incline. You'll need two equations: one for translation, one for rotation.
     
  5. Apr 1, 2005 #4
    So then just switch the sin for cos and the equation is correct?

    (3.4) * (9.8) * (0.5 * 6.8) * cos(48) = 52.4a (where 'a' is angular acceleration)

    Which comes out to 1.446, so to convert this to radians/second you just divide by the radius r=6.8 to get .212742 rads/s^2?



    Also, thanks for the help, Ive solved problem 2 now.
     
    Last edited: Apr 1, 2005
  6. Apr 1, 2005 #5

    Doc Al

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    Right. Realize that "a" (which should be [itex]\alpha[/itex]) is the angular acceleration, which is what you are asked to find. The units of angular acceleration are [itex]{rad}/s^2[/itex].
    No. See my comments above. (No "conversion" needed.)
     
  7. Apr 1, 2005 #6
    Doc Al,

    On an object rolling that does not slip all the frictional force does is allow it to roll, correct? For example, A spool is unwound by pulling at the top, so a force of say 10N is applied by pulling on a string coming unwound at the top, then wouldnt the force of friction where the spool comes in contact with the ground have to be at least 10N?

    Basically Im trying to figure out how to compute force of friction with the ground of a rolling object that does not slip and its not really making sense to me.

    thanks.
     
  8. Apr 1, 2005 #7

    Doc Al

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    From an energy point of view, all the friction does is make it roll with the right speed to avoid slipping. But Newton's 2nd law still applies!

    If the spool moves at a constant speed, then you know the net horizontal force must be zero. In that case you can deduce that the friction force must equal 10N.

    In general, you have to solve for the friction force using Newton's laws. The static friction, a passive force, will be whatever it needs to be to prevent slipping--up to its maximum value of [itex]\mu N[/itex], where N is the normal force between the rolling object and the ground.

    Does this help a little?

    Try solving your second problem using Newton's 2nd law like I mentioned in post #3. What forces act on the ball? What torques do they exert?
     
  9. Apr 1, 2005 #8
    Im sorry to be a pain Doc Al, but you have helped me understand these problems much better.

    [​IMG]


    Lets say F is a constant force of 15N, and the spool will not slip when it rolls.

    When I look at this I want to immediatly jump to the conclusion that if F=15N to the right is applied at the top that on the bottom where the ball contacts the ground that it must be pushing with F=15N to the left there, that much is correct? On top of that you have F=M*g directed straight down? I want to assume that the F_fr must then have to balance these other forces so that F_fr=15N+(m*g), but I do not think this is a correct assumption.

    Again, sorry to be a pain, but you have been more helpful than my book in understanding these concepts, thanks :).
     
  10. Apr 1, 2005 #9

    Doc Al

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    No, this is not correct. The conditions for rolling without slipping will define the relationship between the friction and the applied force. It will turn out that the friction points to the right, the same direction as the applied force. I'll show how to analyze the problem using Newton's 2nd law.

    Note: I'm afraid I may have misled you a bit in my previous answer. When I said that if you pulled the spool with a constant force of 10N to the right, and if the spool moved at a constant speed, then you could deduce that the frictional force must be 10N to the left. This is true. But it would not be rolling without slipping--the spool would slide along the floor; so it would be kinetic friction opposing the motion. So that example does not relate to your problem: how to understand the frictional forces when the cylinder rolls without slipping.

    This is true.

    This is not correct. There are four forces acting on the cylinder, two vertical and two horizontal. The vertical forces add to zero: the weight (mg) acting down is balanced by the normal force of the floor pushing the cylinder up. The horizontal forces do not cancel: as long as there is an applied force, and the cylinder rolls without slipping, it must accelerate.

    Let's analyze those horizontal forces, since they are the only forces that produce rotational or translational acceleration. Let's apply Newton's 2nd law:
    (1) For translational motion: [itex]F + F_{fr} = ma[/itex]
    (2) For rotational motion: [itex](F - F_{fr})R = I\alpha[/itex]

    Be sure you understand where these equation come from. (Note that I assume that the friction force points to the right; if I guessed wrong, the answer will be negative.)

    Now apply two additional facts about this situation:
    (3) Rolling without slipping: [itex]a = \alpha R[/itex]
    (4) Rotational inertia of cylinder: [itex]I = 1/2 m R^2[/itex]

    I'm going to leave the rest to you. Combine these equations to solve for the acceleration and the frictional force.
     
  11. Jan 23, 2011 #10
    hello i need some help in rolling
    i need to know the difference between the static friction and kinetic friction in rolling in an inclined plane and in a horizontal plane what is the right method to study these kinds of problem??
    another problem if someone has a bachelor physics and he wants to continue what options does he have other than teaching
     
  12. Jan 23, 2011 #11

    Doc Al

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    Realize that you are responding to a thread almost 6 years old. It's better to start a new thread.
    If the rolling is without slipping, the friction will be static friction.
    Ask this sort of question in our Career Guidance forum.
     
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