On the picture below you see the motionless black sphere and the green rod rotating around it. At the Position A the rod is motionless, and then it begins rotating and increases this rotational speed up to Position B (you can notice it-the color gradually changes from light green to dark green). After Position B the rod rotates at constant speed I would like to know how I can calculate the acceleration between A and B positions when the rod increases the rotational speed Please Register or Log in to view the hidden image! I know how to compute the acceleration when the rod rotates at the constant speed: a=v^2/r, but what about the situation that I would like to find out? In other words when the rod quickens Please Register or Log in to view the hidden image! let’s assume that length of the rod is 45 meters and it takes 200 seconds to reach Position B from A. So, is there any formula for this purpose? Please Register or Log in to view the hidden image! Uploaded with ImageShack.us
the acceleration from point A to point B isn't constant, it gradualy drops to 0, so it moves in constant speed from point B; so you can only calculate the acceeleration in one point of the rod's trajectory.
The acceleration you describe, a=v²/r, is the acceleration of a mass at the tip of the rod, but it's not the angular acceleration you appear to seek. In rotating systems, θ (theta) is the angle traveled, ω (omega) is the angular velocity, and α (alpha) is angular acceleration. Procedure Measure the radius of the rod Measure the sequential distances between the tips of the rod Use trig to find the angles between the rods (halve the distance, use arcsin function to find the half angle, then double it to find whole angle) Sum the angles in order to find the angle compared to the datum (zero) Assume constant time intervals between angles Using a difference table, insert angles, compute differences to find the angular velocities and then compute the differences of the velocities to find the angular accelerations
Where it is so complicated way? In case of constant acceleration there is simple formula a=v²/r Please Register or Log in to view the hidden image! why such formula does not exist for this case? Please Register or Log in to view the hidden image!
The statement about a=v²/r mixes apples and oranges because r varies along the length of the rod. This equation applies only to a point rotating about a center at a constant velocity. At what distance along the rod's 45-meter radius does this point exist? So, do you want the linear acceleration of a point rotating about a center, or do you want the angular acceleration of a rotating radius? I gave you the angular acceleration of a rotating radius.
Eagle9: The reason that \(a=v^2/r\) can't apply to the whole rod is that \(r\) in that formula is the distance of the accelerating object from the axis of rotation. Different parts of the rod are at different distances from the centre (and are also moving at different speeds). But a bigger problem with that formula is that it gives the centripetal acceleration, which is the part of the acceleration directed towards the centre of the circle. The rod also has a tangential acceleration, which is its acceleration at right angles to the rod. The reason we introduce the concept of angular position, angular acceleration and so on is that every point on the rod always has the same angular acceleration at any given time, whereas different points on the rod have different linear accelerations.
The distance/radius is always the same-45 meters Please Register or Log in to view the hidden image! as for the question for which point are we going to calculate the acceleration: at the end of the rod, in other words at 45 meters away from the black sphere’s center Please Register or Log in to view the hidden image! I know that linear acceleration is calculated by means of this formula: at = (V - Vo)/t and circular with this: ar=V^2/R But how can I “combine” them? That is I want to calculate the circular acceleration when the rotational speed changes, increases Please Register or Log in to view the hidden image!
Eagle9: If you want to know the total acceleration of a point at the end of the rod, we do this: The centripetal acceleration is \(a_c = v^2/r\), where v is the speed of the tip of the rod at any particular time, and r is its distance from the centre. The tangential acceleration is \(a_t = \alpha r\), where \(\alpha\) is the angular acceleration of the rod in radians per second. The total acceleration is the vector sum of the centripetal acceleration and the tangential acceleration. The total acceleration varies as the rod speeds up, because although the tangential acceleration is constant (presumably), the centripetal acceleration increases as the speed increases.
So, how can we add these two kinds of accelerations to each other? I was advised to use Pythagorean Theorem and I think that the following formula is the one that I seek: Please Register or Log in to view the hidden image!
The above is only true for constant angular speed. this is not the case for the OP. While this is true, you can't apply your method since you have the wrong formula for centripetal acceleration. The correct approach is to start with the equations of motion expressed in polar coordinates: \(x=r cos \omega t\) \(y=r sin \omega t\) If you execute the derivatives wrt time and you keep in mind that \(\omega=\omega(t)\) you will get the correct answer. That answer is much more complicated than the one you are suggesting (but it is correct).
Tach: No. My equations are correct. In particular, the centripetal acceleration formula holds for constant or varying velocity.
Eagle9: The magnitude of the total acceleration is: \(|a|=\sqrt{\left(\frac{v^2}{r}\right)^2 + (\alpha r)^2}\)
I don't think so. Start with \(x=r cos \omega t\) \(y=r sin \omega t\) Differentiate once wrt time and you get the speed. Differentiate a second time and you get the acceleration. Only if \( \omega\) is constant do you get:\(a=\frac{v^2}{r}\). For variable \(\omega\) you will not be able to get the above. Here, I'll do the steps for you: \(\frac{dx}{dt}=-r (\omega + \frac{d \omega}{dt} t) sin \omega t \) \(\frac{dy}{dt}=r (\omega + \frac{d \omega}{dt} t) cos \omega t \)
You sure about it? Make \(\frac{d \omega}{dt}=0\) and you recover the well-known: \(\frac{dx}{dt}=-r \omega sin \omega t \) \(\frac{dy}{dt}=r \omega cos \omega t \) \(\frac{d^2x}{dt^2}=-r (\omega)^2 cos \omega t \) Let's see you doing it "right"
Tach: I did it right the first time. You clean up your own errors. I suggest you start by checking your derivatives.
Come on, let's see your math. I am particularly interested in seeing how you get \(a=\frac{v^2}{r}\) for the case when \(\omega\) is variable.
By the way, the full expression for the acceleration of a point in a rotating system is: \(a = a' + \frac{d\omega}{dt} \times r' + 2 \omega \times v' + \omega \times (\omega \times r') + A_0\) where the primed quantities are measured in the rotating reference frame and the unprimed quantities in the non-rotating frame. \(A_0\) is the acceleration of the origin of the primed coordinate frame in the unprimed frame. All quantities are vectors.
