Rolling with or without slipping

NTNUJAVA Virtual Physics Laboratory
Enjoy the fun of physics with simulations!
Backup site http://enjoy.phy.ntnu.edu.tw/ntnujava/

July 07, 2020, 12:09:04 am

Welcome, Guest. Please login or register.
Did you miss your activation email?

Home

Help

Search

Login

Register

"The mind is its own place, and in itself, can make heaven of Hell, and a hell of Heaven." ..."John Milton(1608-1674, English Poet)"

Google Bookmarks

Yahoo My Web

MSN Live

Netscape

Del.icio.us

FURL

Stumble Upon

Delirious

Ask

NTNUJAVA Virtual Physics Laboratory
Enjoy the fun of physics with simulations!
Backup site http://enjoy.phy.ntnu.edu.tw/ntnujava/ > Easy Java Simulations (2001- ) > Ahmed's contribution (EJS simulations) > dynamics (Moderator: ahmedelshfie) > Rolling with or without slipping

Pages: [1] Go Down

« previous next »

	Author	Topic: Rolling with or without slipping (Read 5851 times)
0 Members and 1 Guest are viewing this topic. Click to toggle author information(expand message area).

ahmedelshfie

Moderator
Hero Member

Offline

Offline

Posts: 954

1 Rolling with or without slipping

«

Embed this message

on: April 23, 2010, 01:38:03 am » posted from:,,Brazil

toggle author information

This applet created by Prof Hwang
Modified by Ahmed
Original project Rolling with or without slipping

You can drag both end points of the plane to change the slope (angle)
If the friction coefficient is large enough, the circular object will rolling (without slipping) along the plane. (energy is conserved)
However, if the coefficient is too small, the object will slip along the plane and energy will be loss.
Plotting panel show traces for different energy:

Kinetic energy: (1/2) m v2
rotational energy: (1/2) I w2
potential energy: mgh

Can you identify all the traces?

-*-

You can drag cyan circle inside the rotating object, the trace for this small circle will be shown and the velocity vector(relative to object center ot ground) will be shown,too.

Embed a running copy of this simulation

Embed a running copy link(show simulation in a popuped window)

Full screen applet or Problem viewing java?Add http://www.phy.ntnu.edu.tw/ to exception site list
Press the Alt key and the left mouse button to drag the applet off the browser and onto the desktop. This work is licensed under a Creative Commons Attribution 2.5 Taiwan License

Please feel free to post your ideas about how to use the simulation for better teaching and learning.
Post questions to be asked to help students to think, to explore.
Upload worksheets as attached files to share with more users.

Let's work together. We can help more users understand physics conceptually and enjoy the fun of learning physics!

Rolling with or without slipping.gif (20.4 KB, 765x494 - viewed 552 times.)
	Logged

ahmedelshfie

Moderator
Hero Member

Offline

Offline

Posts: 954

2 Re: Rolling with or without slipping

«

Embed this message

Reply #1 on: April 23, 2010, 01:43:26 am » posted from:,,Brazil

toggle author information

Assume the angle of the slope is $\theta$ , mass of the disk is m, radius is R. The
momentum of inertia is $I=\frac{1}{2}mR^2.$
Let analysis this problem from the contact point between the disk and the slope.
The normal force between the disk and the slope is $mg \cos\theta$ , and the force along the slope is $mg \sin\theta$ ,
Assume the friction between the disk and the slope is f.
The net force is $mg \sin\theta + f = m a$ , where a is the acceleration along the slope.
(the friction force from the slope to the disk is in the same direction as acceleration a)

The condition for rolling without slipping is $a=R\alpha$
The torque is $\tau= R mg\sin\theta = I \alpha =\frac{1}{2}mR^2 \alpha=\frac{1}{2}m R R\alpha=\frac{1}{2}m Ra.$
So $mg\sin\theta=\frac{1}{2}m a, i.e. a =2g \sin\theta and f=ma- mg\sin\theta= mg\sin\theta$
Since the maximum static friction force f_{max}=mg\cos\theta\mu \ge mg\sin\theta,
it imply that $\mu\ge \tan\theta$ for the disk to rolling without slipping.

If it is a ball instead of a disk, then $I=\frac{2}{5}mR^2$ .
$\tau= R mg\sin\theta = I \alpha =\frac{2}{5}mR^2\alpha=\frac{2}{5}m R a$
So $mg\sin\theta=\frac{2}{5}ma, or a=\frac{5}{2}g\sin\theta.$
$f=ma -mg\sin\theta=\frac{3}{5}ma=\frac{3}{2}g\sin\theta$ .
So the condition for rolling without slipping becomes $\mu \ge \frac{2}{3}\tan\theta$ .


« Last Edit: April 23, 2010, 01:45:36 am by ahmedelshfie »	Logged

Pages: [1] Go Up

"The mind is its own place, and in itself, can make heaven of Hell, and a hell of Heaven." ..."John Milton(1608-1674, English Poet)"

« previous next »

Jump to:

Related Topics
		Subject	Started by	Replies	Views	Last post
		Rolling with or without slipping Dynamics	Fu-Kwun Hwang	1	29042	June 11, 2009, 10:24:27 pm by Fu-Kwun Hwang
		Ball Rolling without slipping in a hill Dynamics	Fu-Kwun Hwang	5	27924	March 09, 2009, 11:18:14 pm by Fu-Kwun Hwang
		Bead rolling on hemisphere Question related to Physics or physics related simulation	tanghb	4	13289	March 29, 2015, 04:22:24 am by Femida
		Ball Rolling without slipping in a hill dynamics	ahmedelshfie	0	3839	May 08, 2010, 09:33:43 pm by ahmedelshfie
		Slipping and Rolling Wheel Model Simulations from other web sites	ahmedelshfie	2	8993	September 21, 2010, 11:51:49 pm by ahmedelshfie

Powered by SMF 1.1.13 | SMF © 2006-2011, Simple Machines LLC

Page created in 1.394 seconds with 24 queries.

since 2011/06/15