# When a particle is rotating about an axis "r" away, is its linear velocity constant?

SakuRERE
If talking about a particle rotating around an axis away from it by r. if the particle is moving with constant angular velocity ω. is the linear velocity constant or no?
Now what I know is that since we have Vt= ωr, so r doesn't change, as well as ω, so Vt is said to be constant. but I think it's not right to say that the linear velocity is constant since we have continuous changing in the direction of Vt, but I believe it's a constant linear speed. so, does this mean that at (linear acceleration) is available even when the magnitude of the linear velocity is not changing only the direction does? (since any change in the direction means we have an acceleration)?
when i come to at=α t, if we have an at, then this means we have α which must be zero. so my way of thinking is wrong! right?
this means that it is only for the changing of the magnitude of velocity, right? if yes then what is the linear acceleration that is resulted from the changing of vt direction?

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Mentor
Yes, a particle rotating with a constant angular velocity is always linearly accelerating. Changing the direction of the motion without changing the magnitude of the linear velocity still requires the application of a force.

• SakuRERE
SakuRERE
Yes, a particle rotating with a constant angular velocity is always linearly accelerating. Changing the direction of the motion without changing the magnitude of the linear velocity still requires the application of a force.
but what about at = α r
at is zero and α is zero, how come at is there even when α zero

Homework Helper
but what about at = α r
at is zero and α is zero, how come at is there even when α zero
By "at" do you mean tangential acceleration?

• SakuRERE
In the equations that you have used.

vt = ωr
at = αr

These are not vectors. The left hand side of the first equation is not the velocity. It is the tangential speed. In the second equation, the LHS is the tangential component of the acceleration. It is not the acceleration.
The acceleration, whhich is a vector, has two components. One along the tangent and one along the radius. If the tangential component is zero, which is what happens if ω is constant, then the speed is constant. The velocity is, of course, not constant, because it is changing direction.

• SakuRERE
Cutter Ketch
but what about at = α r
at is zero and α is zero, how come at is there even when α zero

The force and the acceleration are always perpendicular to the current direction of motion, so they only change the direction, not the speed or the kinetic energy.

SakuRERE
y
By "at" do you mean tangential acceleration?
yes yes exactly

SakuRERE
If the tangential component is zero, which is what happens if ω is constant
so the linear acceleration is made of two components :
1- the at the Tangential component, this is related to the magnitude of the velocity only. when there is change in vt there will be at.
2- the radial component which is directed to axis of rotation is related to the direction of the linear velocity only. so when the ω is constant and so vt is constant (α=0 --> at=0) the radial acceleration is still there directed to the axis, since we still have changing in the direction. and the radial acceleration is always there unless the ω=0 means there is no rotation at all! right?
thanks

Yes. You should use subscripts carefully so as not to cause confusion. The quantities are written vt and at.

• SakuRERE
SakuRERE
Yes. You should use subscripts carefully so as not to cause confusion. The quantities are written vt and at.
thanks so much your explanation was straight to the point.
The force and the acceleration are always perpendicular to the current direction of motion, so they only change the direction, not the speed or the kinetic energy.
By "at" do you mean tangential acceleration?
Yes, a particle rotating with a constant angular velocity is always linearly accelerating. Changing the direction of the motion without changing the magnitude of the linear velocity still requires the application of a force.
I am literally appreciating everyone in this forum, who has the willingness to help without any return, every one of you is a lifesaver!

• berkeman