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Motion of a cylinder

  1. Jul 4, 2017 #1
    1. The problem statement, all variables and given/known data

    A right cylinder is at rest as in the figure below. It then starts rolling down without friction slipping. What is the magnitude of its angular velocity when it arrives at point A? At what horizontal distance from A will it stop after hitting the ground?
    Given:
    Moment of inertia of the cylinder ##I = \frac{1}{2}MR^2##.
    Radius of the cylinder ##R = 20 \ cm##.
    Mass of the cylinder ##M = 12 \ kg##.

    ii26B4o.png
    2. The attempt at a solution

    This problem is from an exam from my undergraduate course on physics

    Conservation of energy yields

    Mg6Sin(30°) = MVCM2/2, where VCM is the cylinder center of mass speed and g is the gravity acceleration.

    Since friction isn't present this is a rolling without slipping problem, the linear speed in the equation above should be equal to ωR, where ω is the desired angular speed.

    I got $$\omega = \sqrt{6g} / R = 10 \sqrt{ \frac{3g}{2}} \ \bigg(\frac{ \text{rad}}{ \text{sec}} \bigg)$$

    For the second part of the problem, I considered that the only acceleration is due to gravity (constant), in which case we can use the equations

    V = V0 + at
    D = V0t

    $$0 = V_{0 \ - \ \text{vertical}} + gt \\
    t = |V_{0 \ - \ \text{vertical}}| / |g| = \sqrt{6g} \ \text{Sin(30°)} / g = \frac{1}{2} \sqrt{ \frac{6}{g}} \ (\text{sec})$$

    So, the horizontal distance is $$D = \sqrt{6g} \ \text{Cos(30°)} \times \frac{1}{2} \sqrt{ \frac{6}{g}} = 3 \frac{ \sqrt{3}}{2} \ (\text{meters})$$

    Is this correct?

    EDIT: The correct sentence is "the cylinder is rolling without slipping, instead of without friction".
     
    Last edited: Jul 4, 2017
  2. jcsd
  3. Jul 4, 2017 #2

    SammyS

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    That's a rather strange problem. If there is no friction, then there is no reason for the cylinder to roll. - no reason for its angular momentum to change.
     
  4. Jul 4, 2017 #3
    I'm sorry. I have edited the opening post. Please, consider reading it again.
     
  5. Jul 4, 2017 #4

    SammyS

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    That's more like it.

    That does not include the Kinetic Energy due to the cylinder's rotation.
     
  6. Jul 4, 2017 #5
    Oh, I should have included it... ##I \omega^2 / 2##. So my two answers are incorrect.
     
  7. Jul 4, 2017 #6

    SammyS

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    Yes, they're incorrect.

    Express ##\ v_{CM} \ ## in terms of ω, & solve.
     
  8. Jul 5, 2017 #7
    Thanks.
     
  9. Jul 5, 2017 #8

    SammyS

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    You're welcome.

    Did you get a correct solution ?
     
  10. Jul 5, 2017 #9
    Let me try one solution, this time accounting for the rotation of the cylinder, which takes part of the total energy of it.

    $$Mg \ \text{Sin(30°)} = \frac{1}{2}M \omega^2 R^2 + \frac{1}{4}MR^2 \omega^2 = \frac{3}{4} M \omega^2 R^2 \\
    \Longrightarrow | \omega| = \sqrt{\frac{2}{3}g} \frac{1}{R} = \sqrt{ \frac{50}{3}g} \ \bigg(\frac{ \text{rad}}{ \text{sec}} \bigg) $$
    For the second part of the trajectory,
    $$ | \omega| = \sqrt{ \frac{50}{3}g} \Rightarrow |V_{ \text{CM}}| = \sqrt{ \frac{2}{3}g} \\
    0 = - \sqrt{ \frac{1}{6}g} + gt \Rightarrow t = \frac{1}{ \sqrt{6g}} \ ( \text{sec}) \\
    D = \sqrt{ \frac{2}{3}g} \frac{ \sqrt{3}}{2} \frac{1}{ \sqrt{6g}} = \frac{1}{2 \sqrt{3}} \ ( \text{meters})$$
     
  11. Jul 5, 2017 #10

    SammyS

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    Did you leave out the 6 meters ? Maybe I overlooked that.
     
  12. Jul 5, 2017 #11
    Oh, yes... I'm sorry, I did not remember including it.
    Apart from it, is my reasoning correct?
     
  13. Jul 5, 2017 #12

    SammyS

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    I had not looked in detail at your solution for the second part.

    It definitely is incorrect !

    You seem to be finding the time by thinking that the vertical component of velocity is zero as the cylinder reaches the ground, and furthermore you have gravity working opposite the direction of the velocity. You don't seem to use the fact the the cylinder falls the additional 5 meters after passing point A.
     
  14. Jul 6, 2017 #13
    I do. Note that I'm assuming the vertical component of the velocity to be zero when the cylinder hits the ground and I'm using as initial vertical component the vertical velocity it has when it is at point A. The same reasoning for the horizontal component.
     
  15. Jul 6, 2017 #14

    SammyS

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    That's an incorrect way to solve this.

    It's not the acceleration due to gravity that brings the cylinder to a halt.(The impact with the ground does that.)

    After passing point A, the acceleration due to gravity increases the vertical component of velocity intil the cylinder strikes the ground. There's kinematic equation for that situation from which you can find time.
    Alternatively, you can use conservation of energy to find the speed just prior to impact, and then find distance as you did.​
     
    Last edited: Jul 6, 2017
  16. Jul 7, 2017 #15
    does the distribution of energy between rotating and linear remain constant in proportion regardless of the velocity ?
     
  17. Jul 7, 2017 #16

    haruspex

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    Are you sure that is what the question asks? It will not stop after hitting the ground - it will keep rolling. I guess you mean "will it hit the ground?"
    The SUVAT equations apply only while the acceleration is constant. Hitting the ground involves a very different acceleration, so the equations only apply up to the instant before hitting it.
     
  18. Jul 7, 2017 #17

    haruspex

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    Yes. The ratio does not depend on the radius either.
     
  19. Jul 7, 2017 #18
    You're right. I misstated the question.
    I don't understand it. @SammyS early said that my way of solving the problem is not correct. But it seems that acceleration is constant during the fall. So why can't we use those equations?
     
  20. Jul 7, 2017 #19

    haruspex

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    The acceleration is constant during the fall, but you took the velocity to be zero when it has hit the ground - that is after the fall, not during it.
     
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