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Find the linear acceleration of points on a rolling cylinder

  1. Mar 13, 2017 #1

    vcsharp2003

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    1. The problem statement, all variables and given/known data

    A uniform cylinder with radius r is rolling down an incline without slipping as shown in diagram below. It's angular acceleration is α and acceleration of it's center of mass C is a. What will be the linear acceleration of point of contact P and another point Q on the cylinder ? Assume that points P and Q on the cylinder lie along a circle with center as C.

    Linear_Acceleration_in_Rolling_without_slipping.png

    I have tried to find the linear acceleration by adding linear acceleration vectors resulting from pure rotation and pure translation, but I am not sure if this approach is the right approach in this case. Or whether I am missing something.

    2. Relevant equations
    Linear acceleration resulting from pure rotation for any point on cylinder is given by cross product below.

    $$\overrightarrow{a_t} = \overrightarrow{\alpha} \times \overrightarrow{r} \\ \text { where }\overrightarrow{r} \text { is the radius vector for point in question having a magnitude of } {r}$$

    3. The attempt at a solution
    The motion of cylinder can be considered as having pure rotation defined by angular acceleration as well as pure translation defined by acceleration of center of mass. So linear acceleration of each point will be the vector sum of acceleartions resulting from rotation and translation.

    $$\therefore\overrightarrow{a_p} = \overrightarrow{\alpha} \times \overrightarrow{r_p} + \overrightarrow{a} \\
    \text {and} \\ \overrightarrow{a_q} = \overrightarrow{\alpha} \times \overrightarrow{r_q} + \overrightarrow{a}$$

    Attempt_at_finding_Linear_Acceleration_in_Rollin.png
     
  2. jcsd
  3. Mar 13, 2017 #2
    Hi ,

    The directions of components of acceleration at points P and Q are correct . You could also find magnitude of tangential acceleration (with respect to CM)at the two points .

    Since the cylinder rolls without slipping you also know the net acceleration of the point of contact P. This would give you relation between angular acceleration and acceleration of CM .
     
  4. Mar 13, 2017 #3

    kuruman

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    Can you simplify the expression for ##\vec{a}_P##?
     
  5. Mar 14, 2017 #4

    vcsharp2003

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    Hi,

    $$\overleftarrow{\alpha}\times\overleftarrow{r_p} =|\overleftarrow{\alpha}\rvert \; \rvert\overleftarrow{r_p}\rvert \; \sin90^\circ \text { in an upward tangential direction i.e. parallel to incline and pointing up } \\
    \text {Also , } \rvert\overleftarrow{r_p}\rvert= r \text { where r is radius of cylinder} \\
    \text {Since the tangent at point P is parallel to the incline, so we can simply take} \\ \text { the algebraic sum of two vectors. So, the simplified expression is as below,} \\ \text {if x-axis positive direction is down the plane } \\ \overrightarrow{a_p} =- |\overleftarrow{\alpha}\rvert \; r \; \sin90^\circ\; +|\overleftarrow{a}\rvert
    $$

    You can see the vector diagram for point P, where two accelerations act on point P. The first one is due to angular acceleration and is in the tangential direction at point P having a magnitude of αr and the second acceleration is parallel to the incline plane having a magnitude equal to acceleration of center of mass.
     
    Last edited: Mar 14, 2017
  6. Mar 14, 2017 #5

    vcsharp2003

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    I think I may be missing an acceleration that points towards the center C with a magnitude of v2/r for both points P and Q. This would mean that the linear acceleration of P and Q would be a vector sum of 3 vectors. Does this sound right? This is the centripetal acceleration for a particle moving in a circle.
     
  7. Mar 14, 2017 #6
    I believe you do not have to consider centripetal acceleration in computing the linear accelerations of P and Q .

    Instead if the question had asked to calculate net acceleration of P and Q , then you need to consider centripetal component as well .
     
  8. Mar 14, 2017 #7

    kuruman

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    How are ##\vec{a}## and ##\ | \vec{\alpha }| r \sin90^o## related?

    You also need to be clear whether you are finding the acceleration in the reference frame of the cylinder or of the incline.
     
  9. Mar 14, 2017 #8

    haruspex

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    That is not right.
    Rotational acceleration is only meaningful for a rigid body of some extent. An individual point only has linear acceleration, and that is its net acceleration. In a steadily rotating wheel on a fixed axis, each point is accelerating towards the centre.
     
  10. Mar 14, 2017 #9

    haruspex

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    Yes.
    E.g. consider constant speed on the level. The linear and angular accelerations would be zero, but each point would be accelerating towards the instantaneous centre of rotation.
     
  11. Mar 14, 2017 #10

    vcsharp2003

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    What you said applies to point P where only the centripetal acceleration would be there as other vectors would cancel each other. But for point Q, wouldn't the linear acceleration still be the net of all three accelerations?
     
  12. Mar 14, 2017 #11

    kuruman

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    Suppose I were to define the axis (or centre) of rotation as the set of points on the cylinder that are instantaneously at rest while the rest of the points on the cylinder rotate (also instantaneously) about this axis. Where would you place this axis of rotation? The cylinder rolls without slipping.
     
  13. Mar 14, 2017 #12

    vcsharp2003

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    Not sure about this answer, but I have seen in some books that all the points of contact of cylinder would be instantaneously at rest.
     
  14. Mar 14, 2017 #13

    vcsharp2003

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    I think these two accelerations are equal and opposite if no slipping occurs. I am using the frame of reference as the ground.
     
  15. Mar 14, 2017 #14

    kuruman

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    Consider point R on the incline directly opposite to point P on the cylinder. Rolling without slipping means that P is not moving relative to R. If the entire incline is at rest, what about the points of contact of the cylinder (instantaneously)? Do the books you have seen make sense?
     
  16. Mar 14, 2017 #15

    haruspex

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    The simplest/safest may be to revert to good old Cartesian. Taking axes parallel to and perpendicular to the slope, the centre is at (x, r) and a point on the periphery is at (x+r cos(θ), r+r sin(θ)). Differentiate twice.
     
  17. Mar 14, 2017 #16

    haruspex

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    Right, but that does not help greatly with the accelerations. The point of contact will be accelerating.
     
  18. Mar 14, 2017 #17

    vcsharp2003

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    I do not need the exact answer in terms of x-component or y-component, but just which vectors need to be vectorially added to give the linear accelertion vector.

    So, the point P will have a linear acceleration of v2/r directed towards point C since other two vectors will cancel.
    For point Q, the original vectors shown in OP will be added to a linear acceleration of v2/r directed towards point C, since none of the vectors cancel any other.
     
  19. Mar 14, 2017 #18

    haruspex

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    Yes.
     
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