Getting the magnitude of linear acceleration from angular position equation

AI Thread Summary
The discussion revolves around calculating the magnitude of total linear acceleration from a given angular position equation. The equation provided is \(\theta(t) = \vartheta e^{\beta t}\), with parameters including \(\beta = 2 \, \text{s}^{-1}\) and \(\theta = 1.1 \, \text{rad}\). The user derived angular velocity and acceleration by differentiating the equation twice but received an incorrect linear acceleration result of 2.53 cm/s². Clarification was sought regarding the interpretation of \(\theta = 1.1 \, \text{rad}\), leading to the realization that it should be \(\theta(0) = 1.1 \, \text{rad}\), which may affect the calculations. The conversation highlights the importance of accurate problem definitions in physics.
mconnor92
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Homework Statement


The angular position of an object that rotates about a fixed axis is given by \theta(t) = \vartheta e^\betat,
where \beta= 2 s^−1, \theta = 1.1 rad, and t is in seconds.
What is the magnitude of the total linear acceleration at t = 0 of a point on the object that is 9.2 cm from the axis?
Answer in units of cm/s^2.

Homework Equations


a=\alphar

The Attempt at a Solution


I first plugged in 1.1 for theta and 1/2 for beta since 2^-1 is 1/2. Since the initial equation given is \theta(t) I need to get it into \alpha(t) so that i can plug in t=0 and get the angular acceleration. To do this I took the derivative of the equation twice. After the first derivative I got \omega(t)=.55e^((1/2)t) then after the second I got \alpha(t)=.275e^((1/2)t). I plugged 0 in for t in this equation, giving me \alpha=.275. Then I used the equation a=\alphar to solve for linear acceleration, plugging in .092m for r and .275 for \alpha. This gave me .0253 m/s^2 which I then converted to cm/s^2, resulting in 2.53 cm/s^2. Apparently this is incorrect. Can anyone help me out? Thanks in advance.
 
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mconnor92 said:

Homework Statement


The angular position of an object that rotates about a fixed axis is given by \theta(t) = \vartheta e^\betat,
where \beta= 2 s^−1, \theta = 1.1 rad, and t is in seconds.
What is the magnitude of the total linear acceleration at t = 0 of a point on the object that is 9.2 cm from the axis?
Answer in units of cm/s^2.


Homework Equations


a=\alphar


The Attempt at a Solution


I first plugged in 1.1 for theta and 1/2 for beta since 2^-1 is 1/2. Since the initial equation given is \theta(t) I need to get it into \alpha(t) so that i can plug in t=0 and get the angular acceleration. To do this I took the derivative of the equation twice. After the first derivative I got \omega(t)=.55e^((1/2)t) then after the second I got \alpha(t)=.275e^((1/2)t). I plugged 0 in for t in this equation, giving me \alpha=.275. Then I used the equation a=\alphar to solve for linear acceleration, plugging in .092m for r and .275 for \alpha. This gave me .0253 m/s^2 which I then converted to cm/s^2, resulting in 2.53 cm/s^2. Apparently this is incorrect. Can anyone help me out? Thanks in advance.

Welcome to the PF.

What does it mean by \theta = 1.1 rad ? That will only be true at one moment in time...
 
berkeman said:
Welcome to the PF.

What does it mean by \theta = 1.1 rad ? That will only be true at one moment in time...

Thank you.

I honestly have no idea what \theta = 1.1 rad means. I assumed that it was the value that would be substituted into equation, resulting in \theta(t)=1.1e(1/2)t. Should I then take the derivative from there?
 
mconnor92 said:
Thank you.

I honestly have no idea what \theta = 1.1 rad means. I assumed that it was the value that would be substituted into equation, resulting in \theta(t)=1.1e(1/2)t. Should I then take the derivative from there?

Well, \theta = 1.1 rad cannot be true for all time, since the equation would make no sense then. Does the problem maybe state \theta(0) = 1.1 rad ? That would make more sense, and would give the equation that you are showing...
 
berkeman said:
Well, \theta = 1.1 rad cannot be true for all time, since the equation would make no sense then. Does the problem maybe state \theta(0) = 1.1 rad ? That would make more sense, and would give the equation that you are showing...

But using this assumption gives the same answer as you originally got, which you say is wrong. Maybe the problem definition is missing something?
 
berkeman said:
Well, \theta = 1.1 rad cannot be true for all time, since the equation would make no sense then. Does the problem maybe state \theta(0) = 1.1 rad ? That would make more sense, and would give the equation that you are showing...

Yes it does say that. Sorry I thought I had put that in the original post, but I guess it didn't copy correctly from the homework site. It says \theta0=1.1 and the original equation is \theta(t)=\theta0e\betat
 
berkeman said:
But using this assumption gives the same answer as you originally got, which you say is wrong. Maybe the problem definition is missing something?

I copied and pasted the question directly, so everything given is there. It's possible the system messed up, and I would not be surprised if it did.
 
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