*A ring of radius $R$ rolls on a horizontal ground with linear speed $v$ and angular speed $omega =2v/R$. For what value of $ heta$, the velocity of any point $P$ which makes the given angle with the center, will be in vertically upward direction?*

**My question is:**I know the relation between angular velocity and linear velocity is $v= omega r$.But in the given question it is given that $omega =2v/R$. How is this possible?I assume $v=omega r$ is valid in all situations,but why this discrepancy here?I have seen other such questions as well.

Bạn đang xem: Define angulnr velocity

I asked my friend about it and he told me that both are valid and I am just getting confused.He went on to explain but I couldn"t understand.

Can anybody please explain what I am unable to understand and see?

velocity rotational-kinematics angular-velocity

Share

Cite

Improve this question

Follow

edited Aug 15, 2015 at 18:51

Qmechanic♦

170k3333 gold badges427427 silver badges19611961 bronze badges

asked Aug 15, 2015 at 17:36

Karan SinghKaran Singh

70322 gold badges1212 silver badges2424 bronze badges

$endgroup$

1

Add a comment |

## 2 Answers 2

Sorted by: Reset to default

Highest score (default) Date modified (newest first) Date created (oldest first)

1

$egingroup$

$v = |vecomega imesvecr|= omega r$ is always valid for a rigid rotating body. Here, $r$ refers to the distance of any particular point from a chosen axis of rotation, $omega$, the angular speed of the body about that chosen axis and $v$, the linear speed of that point perpendicular to the radius vector (or the line joining the axis to that point).

In the image given below, for example you could say $$V_perp =|vecomega imesvecr|= omega r = r fracdphidt$$

So it is, in general, a relation between the angular speed and linear speed of a point. This is always true.

However, in the problem something else has been said. In the problem, $v$ refers to the velocity of the center of mass of the ring. So, the problem just says that the center of mass of the body is moving at some particular speed $$v_cm = fracomega R2$$

There is nothing wrong here, the center of mass can move at any speed. Now, if you chose the center of mass as the center of rotation, then you have a rotation at $omega$ superimposed on a translation at $v_cm$. Any rolling motion can be decomposed into a pure rotation and a pure translation, as shown below. Then the two effects can be superimposed.

About this center of mass, $$v_rot=omega r$$ So the point in contact with the ground (at a distance R from the C.M.) for instance, has linear speed $v_cm - Romega$ since rotation and translation oppose each other at that point. As others have pointed out this is not 0 and that implies there is some relative motion between the ring and the ground at the point of contact. This is called slipping.

Xem thêm: Hệ Thống Giáo Dục Lômônôxốp, Trường Thcs Và Thpt, Hệ Thống Giáo Dục Lômônôxốp, Trường Thcs Và Thpt

So your friend was correct. Both are valid and correct. But they refer to different things - one to the motion of the center of mass and the other to that of any point on the body.