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 Periodic and Circular Motion Periodic and Circular Motion Physics Help Forum May 18th 2015, 04:19 PM #1 Junior Member   Join Date: May 2015 Posts: 5 Rotational accelerations I am going through a practice question in which a disc of radius 3.5cm is rotating, it makes one revolution every second. I have to work out the tangential speed, tangential acceleration and angular acceleration. I can work out the tangential speed but am struggling with the accelerations. Would the tangential acceleration just be the tangential speed/time and angular acceleration be centripetal acceleration? Thanks   May 18th 2015, 05:06 PM   #2
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 I am going through a practice question in which a disc of radius 3.5cm is rotating, it makes one revolution every second. I have to work out the tangential speed, tangential acceleration and angular acceleration. I can work out the tangential speed but am struggling with the accelerations. Would the tangential acceleration just be the tangential speed/time and angular acceleration be centripetal acceleration? Thanks
Centripetal force is the net force on an object, usually when moving in a circle at constant speed. The centripetal acceleration always points to the center of the circle so it can't be the tangential acceleration nor can it be the angular acceleration. (Note also that centripetal acceleration has the units m/s^2, not rad/^2.)

As you say, s = r(theta), v = r(omega), and a = r(alpha), where r = 3.5 cm and s, v, and a are tangential values.

(omega) is given as 1 rev/s = (2 pi ) rad/s... v = r(omega). From the problem statement it looks like (omega) is constant, so (alpha) is 0 rad/s^2, thus a = r(alpha) = 0 cm/s^2.

Until you get used to the idea of rotational mechanics I'd recommend a cheat sheet in front of you whenever you do one of these problems. It's no harder than the usual kinematics you've already studied but for some reason the Greek variables throw everyone off.

-Dan
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See the forum rules here.   May 19th 2015, 01:09 AM   #3
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 I am going through a practice question in which a disc of radius 3.5cm is rotating, it makes one revolution every second. I have to work out the tangential speed, tangential acceleration and angular acceleration. I can work out the tangential speed but am struggling with the accelerations. Would the tangential acceleration just be the tangential speed/time and angular acceleration be centripetal acceleration? Thanks
No the angular acceleration is not the centripetal acceleration.

Acceleration of any kind is rate of change of something.

Tangential acceleration is rate of change of tangential velocity.

Angular acceleration is rate of change of angular velocity.

So if your rotating body is rotating at constant angular velocity, it's angular acceleration is zero.

Since the tangential velocity = angular velocity x radius x 2pi, constant angular velocity makes for constant tangential velocity.

Now acceleration, like velocity is a vector quantity, which means it can be resolved into components in mutually perpendicular directions.

In the case of rotation we consider radial and tangential directions as the mutually perpendicular directions.

When a body is rotating at constant angular velocity, with no external forces acting;

The radial velocity is zero
The radial acceleration is called the centripetal acceleration and is non zero

The tangential velocity is (2pi)(R)(omega) as already noted
The tangential acceleration is zero

Are you happy so far?

Last edited by studiot; May 19th 2015 at 01:18 AM.   May 20th 2015, 09:35 AM #4 Senior Member   Join Date: Apr 2008 Location: Bedford, England Posts: 668 Do we have to consider the reference frame for the velocity and acceleration? Viewed from within the rotating reference frame of the disc, the tangential velocity is constant and the acceleration is zero. However; Viewed from a non-rotating reference frame, the velocity has constant magnitude, but a changing direction. Does this not also imply that it has an acceleration... __________________ You have GOT to Laugh !   May 20th 2015, 12:16 PM   #5
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 Does this not also imply that it has an acceleration...
Yes, there is definitely acceleration of a point at the edge of the disc, but that acceleration is in the radial direction, not the tangential direction (as long as both the rotational velocity and the radius of rotation are constant, as they appear to be in this case). The equations for radial acceleration (a_r) and tangential acceleration (a_theta) for a rotating frame are: where "theta hat" and "r hat" are unit vectors in the tangential and radial directions respectively. You can see that if rotational acceleration (theta double dot) is zero and r is constant then the only component of acceleration is in the radial direction (the "r hat" direction).

Last edited by ChipB; May 20th 2015 at 01:26 PM.   May 20th 2015, 03:56 PM #6 Senior Member   Join Date: Apr 2008 Location: Bedford, England Posts: 668 Ok, I was over thinking it. As often seems to be the case with problems posed for teaching purposes, they have added in the questions about accelerations almost as read herrings. One is moved to think that the straightforward solution (they are both zero) is too easy, and then you try to think of a reason why this solution might not be correct. I suppose also that the fact that the question asks for tangential and angular velocities and accelerations places the problem firmly in the rotational frame. __________________ You have GOT to Laugh !   May 20th 2015, 04:01 PM   #7
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 One is moved to think that the straightforward solution (they are both zero) is too easy, and then you try to think of a reason why this solution might not be correct.
Just to be clear, radial acceleration is NOT zero for this problem.   Sep 8th 2016, 06:32 PM #8 Senior Member   Join Date: Aug 2010 Posts: 434 I did not know herring could read!   Sep 27th 2016, 11:52 PM #9 Junior Member   Join Date: Sep 2016 Location: London Posts: 21 Angular acceleration(Rotational accelerations) Angular velocity is the rate of change of orientation of a body. Angular acceleration is the rate of change of angular velocity. Angular velocity is the limiting time rate of change of angular displacement. ω = ∆ø/ ∆t In the aforementioned formula, ω stands for angular velocity, ø stands for angular displacement and t represents time. In case a particle performing uniform circular motion completes one rotation, its angular velocity will be equal to 2π/t since angular displacement will be equal to 2π ( circumference of a circle). Angular acceleration is the rate of change of angular velocity. α =∆ω/∆t in this formula, angular acceleration (α) is equal to change in angular velocity divided by time.  Tags accelerations, rotational Search tags for this page

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