# Is the acceleration of a rotating body always it's centripetal component? [closed]

For constant circular motion where a rotating mass accelerates angularly. Would the linear acceleration of the rotating mass always be equal to its centripetal acceleration

e.g. The earth's centripetal acceleration around the Sun is $$0.006m/s^2$$ does this mean one can say the earth's acceleration when it travels(revolves) at $$670,000mph$$ is simply equal to its centripetal acceleration?

• I don't understand your question. Velocity is a vector. Change in velocity is acceleration, which is also a vector. Jan 5 '19 at 15:16
• Possible duplicate of Centripetal force for non-uniform circular motion Jan 5 '19 at 15:32
• @Mike Dunlavey for angular velocity we have that $$\omega = v/r$$ which is a function of linear velocity(v) over the radius(r). The centripetal acceleration is $$a_{cp} = \omega ^2r$$ and the angular acceleration is $$\omega /t$$ the question is if the centripetal $(a_{cp})$ acceleration is always equal to the linear acceleration $(a)$ which is a function of the linear velocity $(v)$ Jan 5 '19 at 15:36
• @LiNKeR: Centripetal acceleration is also $v^2/r$, so I think the answer is yes. By the way, $\omega/t$ is the angular velocity, not acceleration, and your figure for earth's speed in orbit is too large by a factor of 10. Jan 6 '19 at 14:20

Yes, because the force is also centripetal. This is only for circular motion at constant speed, if it is rotating at changing speed there will be also a component of the acceleration tangential to the circle, and the resultant linear acceleration will no longer be centripetal (that is, radial)

• would it be correct if the speed is not constant to then have that $$a_{cp} = \omega ^2r \\ \text{ where } \omega \text{ is the angular speed } \\ a_{cp} = \omega (\omega r) \\ \text{ since } v = \omega r \\ a_{cp} = \omega v \\ \text{ also } v = at \\ a_{cp} = \omega (at) \\ \text{ from } \omega = \theta/t \\ a_{cp} = \omega t(a) \\ a_{cp} = \theta (a) \\ a = a_{cp}/\theta$$ where $a$ is the linear acceleration and $a_{cp}$ is the centripetal acceleration Jan 5 '19 at 15:40
• no, v=at only in the uniform acceleration case (linear motion), in the uniform rotating case the acceleration is not uniform. It is uniform in magnitude, but not in direction.
– user65081
Jan 5 '19 at 15:45
• @Wolfram jonny thanks for clearing the confusion a little but are you saying if I spin a mass attached to a rope while increasing angular speed by $25°$ each time the angular acceleration would not be constant or its just the linear acceleration that would not be Jan 5 '19 at 16:06
• if by "a" you mean the uniform angular acceleration, then $\omega=\theta/t$ is wrong, because now $\theta=\omega t +1/2 a t^2$
– user65081
Jan 5 '19 at 16:16
• I meant $a$ to be linear acceleration $\omega = \theta /t$ => angular speed = angle/time Jan 5 '19 at 16:30

Acceleration is a vector quantity defined as $$\Delta v/\Delta t$$ for discrete time intervals. Velocity is also a vector quantity, defined as $$\Delta x/\Delta t$$ for discrete time intervals, where $$\Delta x$$ denotes displacement.

For an acceleration to occur, velocity must be changing. There are two ways that this can happen. For an object moving in circular motion at constant speed, the velocity is continuously changing because the direction of motion is continuously changing. Thus, even though the speed is constant, the object is accelerating, and this case of acceleration is what is known as centripetal acceleration, with the associated equation $$a_c=v^2/r$$.

For an object traveling in a straight line, the velocity can change by changing the speed of the object. This is the more "familiar" example of acceleration, known as linear acceleration, with the associated equation $$a=\Delta v/\Delta t$$.

Note that these two equations have different functional forms, and they are NOT the same. In other words, centripetal acceleration should NOT be equated with linear acceleration.