Question

Rolling motion:

how will the velocity of a point on the sides of the object differ from velocity of COM in an ideal rolling motion (no skipping/sliding)?

Answer 1

Why is D not true?

Answer 2

So we have two different questions:
For the first:
1) What is the relationship between v tangential, r and omega?
For the second:
what is the relationship between a tangential, r and alpha?
what is the relationship between a radial, omega and r?

Answer 3

and all these are in ideal rolling conditions.

Answer 4

i know the general formulas and how they apply. its just confusing why the vertical component of the velocity at the side is the same as the velocity of COM.

Answer 5

Try drawing a diagram...and label your velocity and acceleration vectors.

Answer 6

i understood the reason for the velocities being similar. Now, why does the acceleration at top not equal 0 when an object has constant linear velocity. (im assuming that omega wouldn't change so by the formula atop=0+2(0)r=0)

Answer 7

Because velocity still changes in direction.
The radius of curvature of the path is never infinite in this example, hence a = v²/r

Answer 8

I can't read anything after "a=" but i get what you're trying to say. So, is alpha non zero then? Also, does it matter for tangential acceleration? (since i learnt that an object has tangential acceleration only when there is a change in 'angular' velocity)

Answer 9

If omega is constant, what is alpha equal to?

Answer 10

0

Answer 11

ok, so if alpha is constant , the tangential acceleration is zero since:
a(tangential) = alpha*r ....agreed

Answer 12

I meant if alpha = 0

Answer 13

agreed.

Answer 14

so, why does the book say otherwise?

Answer 15

hang on...

Answer 16

But there is still a radial acceleration since omega is constant.
That is, a radial = v^2/r = omega^2*r
So for a point on the edge of the tire there is no tangential acceleration but there is a radial acceleration.

Answer 17

oh! Thanks for clearing that up. (i didn't realize they meant both types)

Answer 18

Excellent! It is very important to remember that when you are dealing with circular motion you need to consider both possible accelerations...a tangential which depends on alpha and a radial which depends on omega

Answer 19

Thanks to @vincent-lyon.Fr as well!

Answer 20

correct me if im wrong: radial acceleration can be thought of as a measure of the acceleration due to change in direction and tangential as a measure of acceleration due to change in angular velocity.

Answer 21

Yes