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Consider the figure:enter image description here

My text book says:

As the body rotates, the point P moves along a circle of radius r with a linear velocity v whereas the line OP rotates with angular velocity w as shown in fig5.4(b)

Question is that doesn't the point P(being moves in a circular path around a fixed axis) also have an angular velocity just like line OP? Moreover, I couldn't understand why the OP doesn't possess linear velocity v in addition to angular velocity?

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  • $\begingroup$ I think the book is implying that the apoapsis is precessing, thus the orbital path can be said to have an angular momentum. Yes, the orbiting body has orbital angular momentum as well, but that is not what the question seems to be concerned with. $\endgroup$ – honeste_vivere May 22 '17 at 14:26
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The book is a little bit confusing - but both it and you are correct.

Yes, the point P describes a circular path; this means that at any moment it has a velocity $v=\omega r$.

A point doesn't really have "angular velocity" by itself; you can't tell it is rotating because it has no angular momentum about its center of mass. It does require a constant (centripetal) acceleration to stay in the circular path.

The entire line OP, on the other hand, can be seen to be rotating: if it has a mass per unit length, the total energy would be greater than the kinetic energy of its center of mass (which is moving at $v=\frac12\omega r$). Every point on the line has a different velocity (because it has a different distance $r$ to the center of rotation). So it doesn't make as much sense to speak of the velocity of OP.

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  • $\begingroup$ ,if point p is not rotating then why it follows a circular trajectory?and how we come to know that it has no angular momentum about its center of mass as no values or any data is mentioned; after all it is a theory and not numerical.? $\endgroup$ – M.Naeem Ahmad May 22 '17 at 15:12
  • $\begingroup$ A point has no size and therefore no angular momentum. It follows a circular trajectory because it is attached to a body that does have finite size (and therefore angular momentum). $\endgroup$ – Floris May 22 '17 at 15:13
  • $\begingroup$ ,if point P has no size then it should also not possess linear momentum, (and so no linear velocity)but it has linear velocity....why? $\endgroup$ – M.Naeem Ahmad May 22 '17 at 15:24
  • $\begingroup$ A point can have mass without having size. $\endgroup$ – Floris May 22 '17 at 15:26
  • $\begingroup$ ,so could we say that linear momentum (and velocity) is related to body's mass....while angular momentum (and angular velocity) is related to particle's size,isn't it? $\endgroup$ – M.Naeem Ahmad May 22 '17 at 15:45
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Only solids can have angular velocity, by definition. Forget the layman's definition of a solid, a solid in mathematics is as follows:

Consider R3 space, where every point has a position vector with 3 coordinates. This in itself is a "fixed" solid, a solid that isn't moving. A solid is R3 space such that no point is moving with respect to another. In traditional R3 space, all points are fixed, hence they aren't moving with respect to each other.

Now keep this fixed R3 space in your mind. We define another R3 space that is superposed onto the fixed R3 space, such that every point in fixed R3 space corresponds to another point in the new R3 space. The difference is that this new R3 space is moving, but it is moving TOGETHER, such that within the R3 space there is no relative movement of its points between each other. That is a solid, and its angular velocity is the angular rate of change of the solid around its axis of rotation.

Hence, when a point is moving AS PART OF A SOLID with a specific angular velocity, we say that the point has that angular velocity. If its movement does not correspond to a movement of this nature, then one cannot define its angular velocity.

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The question is a little confusing but let me try to put some information in a way that hopefully will answer the question. I assume that the line is a solid wore or a thin rod.

In such a case every particle on the rod will posses linear velocity $v$ that reduces as you move from point P towards the center. The center actually has zero linear velocity (visualize the wing of a fan, you can trace the movement of inner edge of the wing but not the outer. The reason is that the velocity of outer edge is more than that of the inner hence difficult to track by eyes). The other way of interpreting is that the outer edge has to cover a larger circumference (owing to larger radius) than the inner edge (smaller radius) in same time $t$. Thus center that has zero radius has zero velocity.

However, the angular velocity $\omega$ measures the angular sweep of the wire OP per unit time. You can clearly see that every point on OP sweeps a similar angle and hence the angular velocity of every point on OP is the same.

So the angular velocity of the wire and its every point is same but the linear velocity reduces as you move from P towards O.

The relation between linear velocity v and angular velocity is given by the equation

$v = \omega r$

Mathematically also since $\omega$ is constant, $v$ must increase as $r$ increases and reduce as $r$ reduces

Use this note to figure out any answer around rotational motion of OP

You may like to watch this video on the topic made by me for better understanding

Rotation - Angular displacement, Velocity and acceleration

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  • $\begingroup$ Thanks! How do you do this change? $\endgroup$ – Vish Jan 30 '18 at 4:30

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