How Is Acceleration Calculated in Rotational Motion?

In summary, the conversation discusses the calculation of tangential and normal acceleration using formulas related to angular velocity and acceleration, as well as the vector sum of these two components. There were some errors in the numerical calculations, but the basic concept was correct. It was also noted that the term "linear" acceleration was confusing in the original problem statement.
  • #1
Vladimir_Kitanov
44
14
Homework Statement
On picture
Relevant Equations
##\theta = \theta_0 + \omega _0 t + \frac {1}{2} \alpha t^2##
##\omega = \omega_0 + \alpha t##
##\omega^2 = \omega_0^2 + 2 \alpha (\theta - \theta_0)##
##a_t = \alpha * r##
##a_n = \omega^2 * r##
Capture.png

Answer should be (c) 32,7 ##\frac{m}{s^2}##

My attempt:

##\omega_{2\pi}## -> ##\omega## after 1 revolution
##\omega_{2\pi} = 0,2 * (2\pi)^2##
##\omega_{2\pi} = 7,9 \frac{rad}{s}##

##\frac {d}{dt}\omega = \alpha = 0,2*2*\theta##
##\alpha_{2\pi}## -> ##\alpha## after 1 revolution
##\alpha_{2\pi} = 2,51 \frac {rad}{s}##

##a_n## -> normal acceleration
##a_t## -> tangential acceleration

##a_t = \alpha_{2\pi} * r = 1,26 \frac {m}{s^2}##
##a_n = \omega^2 * r = 32,205 \frac {m}{s^2}##
##a = \sqrt{a_n^2 + a_t^2} = 31,2 \frac{m}{s^2}##
 
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  • #2
Vladimir_Kitanov said:
##\frac {d}{dt}\omega = \alpha = 0,2*2*\theta##
##\alpha_{2\pi}## -> ##\alpha## after 1 revolution
##\alpha_{2\pi} = 2,51 \frac {rad}{s}##
Does ##\alpha## have the proper units of angular acceleration in your derivation?
 
  • #3
Vladimir_Kitanov said:
##\frac {d}{dt}\omega = \alpha = 0,2*2*\theta##
I am concerned with:
##\frac {d}{dt}\omega = \alpha = 0,2*2*\theta##

You are given ##\omega## as a function of ##\theta##. But you appear to have differentiated with respect to ##\theta## and obtained a derivative with respect to ##t##.
 
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  • #4
This might be easier to think about the differentiation
## \omega = 0.2 \cdot (\theta(t))^2##

Note, some of the formulas you listed as "relevant" does not apply here. Why is it so?
 
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  • #5
malawi_glenn said:
Note, some of the formulas you listed as "relevant" does not apply here. Why is it so?
I just put them there.
They are useful in some problems like that.
 
  • #6
erobz said:
Does ##\alpha## have the proper units of angular acceleration in your derivation?
No ##\alpha## have unit ##\frac{rad}{s^2}##
But that does not change anything.
 
  • #7
jbriggs444 said:
I am concerned with:
##\frac {d}{dt}\omega = \alpha = 0,2*2*\theta##

You are given ##\omega## as a function of ##\theta##. But you appear to have differentiated with respect to ##\theta## and obtained a derivative with respect to ##t##.
That is my mistake.
Thanks.

It should be:
##\alpha = 0,2*2*\theta*\omega##
Derivative of ##\theta## with respect to time is ##\omega##
 
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  • #8
Vladimir_Kitanov said:
No ##\alpha## have unit ##\frac{rad}{s^2}##
But that does not change anything.
But it does.

For future reference: If you derive something, and its units do not match the quantity you are after then there is a mistake in said derivation.
 
  • #9
To summarize, $$\alpha=\frac{d\omega}{dt}=\frac{d\omega}{d\theta}\frac{d\theta}{dt}=\omega \frac{d\omega}{d\theta}.$$What is the tangential component ##a_t##?
 
  • #10
kuruman said:
$$\alpha=\frac{d\omega}{dt}=\frac{d\omega}{d\theta}\frac{d\theta}{dt}=\omega \frac{d\omega}{d\theta}.$$What is the tangential component ##a_t##?
##a_t = \alpha * r##
##\alpha## is angular acceleration.
Capture.png
 
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  • #11
The term “linear” acceleration is confusing in the problem statement. What I think is being calculated is just the magnitude of the acceleration?
 
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  • #12
erobz said:
The term “linear” acceleration is confusing in the problem statement. What I think is being calculated is just the magnitude of the acceleration?
Yes
 
  • #13
Vladimir_Kitanov said:
Yes
You got the basic concept correct in your OP but just are having issues with the numerical calculations. It is the vector sum as you tried to compute originally. You just need to be very careful computing ##a_n## and ##a_t## especially the part explained by @kuruman in post #9.
 
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  • #14
Vladimir_Kitanov said:
##a_t = \alpha_{2\pi} * r = 1,26 \frac {m}{s^2}##
##a_n = \omega^2 * r = 32,205 \frac {m}{s^2}##
##a = \sqrt{a_n^2 + a_t^2} = 31,2 \frac{m}{s^2}##
In addition to the other comments, can I add that the hypotenuse is the longest side of a right-triangle.

(Minor edit.)
 
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FAQ: How Is Acceleration Calculated in Rotational Motion?

What is rotation about a fixed axis?

Rotation about a fixed axis is a type of rotational motion in which an object rotates around a single, fixed axis. This means that the object's axis of rotation remains constant throughout the motion.

What is the difference between rotation about a fixed axis and rotation about a moving axis?

The main difference between rotation about a fixed axis and rotation about a moving axis is that in the former, the axis of rotation remains constant while in the latter, the axis of rotation changes over time.

What is the relationship between angular velocity and linear velocity in rotation about a fixed axis?

In rotation about a fixed axis, the angular velocity (ω) and linear velocity (v) are related by the equation v = ωr, where r is the distance from the axis of rotation to the point on the object. This means that the linear velocity is directly proportional to the angular velocity and the distance from the axis of rotation.

What is moment of inertia in rotation about a fixed axis?

Moment of inertia is a measure of an object's resistance to rotational motion. In rotation about a fixed axis, it is calculated by summing the products of each particle's mass and its distance from the axis of rotation squared.

What is the conservation of angular momentum in rotation about a fixed axis?

The conservation of angular momentum states that in a closed system, the total angular momentum remains constant. In rotation about a fixed axis, this means that if no external torque is applied, the angular momentum of the object will remain constant.

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