How Do You Calculate the Motion of a Pebble in a Rolling Tire?

  • Thread starter Thread starter HeLLz aNgeL
  • Start date Start date
  • Tags Tags
    Rolling Tire
Click For Summary
SUMMARY

The discussion focuses on calculating the motion of a pebble embedded in the tread of a rolling tire with an outer radius R and angular velocity omega. The participants derive the velocity vector of the axle relative to the ground as v_vec_ag(t) = -Rw x_unit. They also explore the position vector of the pebble, r_pg_vec(t), which accounts for both the rotation and translation of the tire, resulting in r_pg_vec(t) = (-Rsin(ωt) - Rωt)x_unit + (Rcos(ωt) + R)y_unit. The velocity vector of the pebble, v_vec_pg(t), and the acceleration vector, a_vec_pg(t), are subsequently derived through differentiation.

PREREQUISITES
  • Understanding of angular motion and angular velocity
  • Familiarity with vector calculus and unit vectors
  • Knowledge of trigonometric functions (sine and cosine)
  • Basic principles of rolling motion and kinematics
NEXT STEPS
  • Study the derivation of position vectors in rotational motion
  • Learn about the relationship between linear and angular velocity
  • Explore the concepts of differentiation in vector calculus
  • Investigate the effects of rolling without slipping in physics
USEFUL FOR

Students in physics, particularly those studying mechanics, engineers working with rotational systems, and anyone interested in the dynamics of rolling objects.

HeLLz aNgeL
Messages
30
Reaction score
0

Homework Statement


You are to find the coordinates of a pebble stuck in the tread of a rolling tire that is rotating counterclockwise (i.e., in the positive sense) with angular velocity omega. The tire rolls without slipping on the ground (which is at y = 0 ). The outer radius of the tire is R. At time t = 0 , the pebble is at the top of the tire, as shown.

a). Find the velocity of the axle of the tire relative to a fixed point on the ground, v_vec_ag(t). Note the order of the subscripts: velocity of axle measured relative to the ground. Express your answer in terms of R, omega, and x_unit and/or y_unit.

The pebble and tire have now rolled as shown in the figure. View Figure Answer the following questions for t>0.

b). Find the position vector of the pebble relative to the initial point of contact between the wheel and ground at a time t, r_pg_vec(t).
Express the position vector of the pebble in terms of R, omega, t, and the unit vectors x_unit and/or y_unit of the xy coordinate system shown.


c). Find v_vec_pg(t), the velocity vector of the pebble with respect to a fixed point on the ground, in terms of the unit vectors x_unit and y_unit of the xy coordinate system shown.
Express the velocity vector in terms of R, omega, t, and x_unit and/or y_unit.


d) Now find a_vec_pg(t), the acceleration vector of the pebble with respect to a fixed point on the ground.
Express your answer in terms of R, omega, t and x_unit and/or y_unit of the xy coordinate system shown.

The Attempt at a Solution



ok, for the first one, i know v = rw, but how do i write it in terms of vectors, v =rwx ?

and for part b, i have no clue how i should be approaching this ... just seems too weird :S
 

Attachments

  • 1.jpg
    1.jpg
    7.2 KB · Views: 1,069
  • 2.jpg
    2.jpg
    7.7 KB · Views: 1,021
Physics news on Phys.org
I think for part a), what they want is:

v_vec_ag(t) = -Rw\hat{x}... ie -Rw times a unit vector in the x-direction.

For part b), suppose the wheel was just rotating without any translation... (ie it is just rotating in place). What is the position vector relative to the initial point of contact? use sin, cos etc... the angle through which the wheel has rotated is wt...

then how does this position vector change when you take into account the horizontal velocity of the tire?

c) take the derivative of b).

d) take the derivative of c).
 
ok, so i know its Rsinwt(x) + RCoswt(y) and the initial position was just R(y)

so, relative to the initial position it should be Rsinwt(x)+(R(y)-RCoswt(y))

doesnt sound right, does it ?
 
bump ... anyone ?
 
HeLLz aNgeL said:
ok, so i know its Rsinwt(x) + RCoswt(y) and the initial position was just R(y)

so, relative to the initial position it should be Rsinwt(x)+(R(y)-RCoswt(y))

doesnt sound right, does it ?

Well... if translation wasn't there... here's what I get:

\vec{r} = -Rsin(wt)\hat{x} + (Rcos(wt)+R)\hat{y}

Now add the translation... -Rwt\hat{x}

so it comes out to:

\vec{r} = (-Rsin(wt)-Rwt)\hat{x} + (Rcos(wt)+R)\hat{y}
 
Last edited:

Similar threads

  • · Replies 42 ·
2
Replies
42
Views
3K
  • · Replies 11 ·
Replies
11
Views
16K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
1
Views
8K
Replies
11
Views
7K
Replies
1
Views
12K
  • · Replies 14 ·
Replies
14
Views
2K
Replies
40
Views
2K
  • · Replies 60 ·
3
Replies
60
Views
4K
  • · Replies 5 ·
Replies
5
Views
1K