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/

May 22, 2018, 07:56:43 am

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

Home

Help

Search

Login

Register

Progress is impossible without change; and those who cannot change their minds cannot change anything. ..."George Bernard Shaw(1865-1950, Irish dramatist, essayist and critic, Nobel Prize for Literature 1925)"

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- ) > Dynamics > Rolling with or without slipping

Pages: [1] Go Down

« previous next »

	Author	Topic: Rolling with or without slipping (Read 25191 times)
0 Members and 1 Guest are viewing this topic. Click to toggle author information(expand message area).

Fu-Kwun Hwang

Administrator
Hero Member

Offline

Offline

Posts: 3080

WWW

1 Rolling with or without slipping

«

Embed this message

on: July 05, 2008, 12:45:07 pm »

toggle author information

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!


	Logged

Fu-Kwun Hwang

Administrator
Hero Member

Offline

Offline

Posts: 3080

WWW

2 Re: Rolling with or without slipping

«

Embed this message

Reply #1 on: June 11, 2009, 10:24:27 pm » posted from:Taipei,T\'ai-pei,Taiwan

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 $\mu \ge \frac{2}{3}\tan\theta$ .


	Logged

Pages: [1] Go Up

Progress is impossible without change; and those who cannot change their minds cannot change anything. ..."George Bernard Shaw(1865-1950, Irish dramatist, essayist and critic, Nobel Prize for Literature 1925)"

« previous next »

Jump to:

Related Topics
		Subject	Started by	Replies	Views	Last post
		Ball Rolling without slipping in a hill Dynamics	Fu-Kwun Hwang	5	25562	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	11521	March 29, 2015, 04:22:24 am by Femida
		Rolling with or without slipping dynamics	ahmedelshfie	1	5077	April 23, 2010, 01:43:26 am by ahmedelshfie
		Ball Rolling without slipping in a hill dynamics	ahmedelshfie	0	3343	May 08, 2010, 09:33:43 pm by ahmedelshfie
		Slipping and Rolling Wheel Model Simulations from other web sites	ahmedelshfie	2	7983	September 21, 2010, 11:51:49 pm by ahmedelshfie

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

Page created in 1.407 seconds with 22 queries.

since 2011/06/15