Getting the magnitude of linear acceleration from angular position equation

In summary: I'll try contacting my professor. Thank you for your help!In summary, the problem involves finding the total linear acceleration of an object at t=0 given its position and angular velocity. The equation \theta(t) = \theta0e\betat is provided, where \theta0 = 1.1 rad and \beta = 2 s^-1. The steps to solve involve taking the second derivative of the equation and using the equation a=\alphar to find the linear acceleration. However, despite following these steps, the given answer of 2.53 cm/s^2 does not match the calculated result. It is possible that there is missing information or a mistake in the problem definition.
  • #1
mconnor92
4
0

Homework Statement


The angular position of an object that rotates about a fixed axis is given by [tex]\theta[/tex](t) = [tex]\vartheta[/tex] e^[tex]\beta[/tex]t,
where [tex]\beta[/tex]= 2 s^−1, [tex]\theta[/tex] = 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=[tex]\alpha[/tex]r

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 [tex]\theta[/tex](t) I need to get it into [tex]\alpha[/tex](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 [tex]\omega[/tex](t)=.55e^((1/2)t) then after the second I got [tex]\alpha[/tex](t)=.275e^((1/2)t). I plugged 0 in for t in this equation, giving me [tex]\alpha[/tex]=.275. Then I used the equation a=[tex]\alpha[/tex]r to solve for linear acceleration, plugging in .092m for r and .275 for [tex]\alpha[/tex]. 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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  • #2
mconnor92 said:

Homework Statement


The angular position of an object that rotates about a fixed axis is given by [tex]\theta[/tex](t) = [tex]\vartheta[/tex] e^[tex]\beta[/tex]t,
where [tex]\beta[/tex]= 2 s^−1, [tex]\theta[/tex] = 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=[tex]\alpha[/tex]r


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 [tex]\theta[/tex](t) I need to get it into [tex]\alpha[/tex](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 [tex]\omega[/tex](t)=.55e^((1/2)t) then after the second I got [tex]\alpha[/tex](t)=.275e^((1/2)t). I plugged 0 in for t in this equation, giving me [tex]\alpha[/tex]=.275. Then I used the equation a=[tex]\alpha[/tex]r to solve for linear acceleration, plugging in .092m for r and .275 for [tex]\alpha[/tex]. 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 [tex]\theta[/tex] = 1.1 rad ? That will only be true at one moment in time...
 
  • #3
berkeman said:
Welcome to the PF.

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

Thank you.

I honestly have no idea what [tex]\theta[/tex] = 1.1 rad means. I assumed that it was the value that would be substituted into equation, resulting in [tex]\theta[/tex](t)=1.1e(1/2)t. Should I then take the derivative from there?
 
  • #4
mconnor92 said:
Thank you.

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

Well, [tex]\theta[/tex] = 1.1 rad cannot be true for all time, since the equation would make no sense then. Does the problem maybe state [tex]\theta(0)[/tex] = 1.1 rad ? That would make more sense, and would give the equation that you are showing...
 
  • #5
berkeman said:
Well, [tex]\theta[/tex] = 1.1 rad cannot be true for all time, since the equation would make no sense then. Does the problem maybe state [tex]\theta(0)[/tex] = 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?
 
  • #6
berkeman said:
Well, [tex]\theta[/tex] = 1.1 rad cannot be true for all time, since the equation would make no sense then. Does the problem maybe state [tex]\theta(0)[/tex] = 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 [tex]\theta[/tex]0=1.1 and the original equation is [tex]\theta[/tex](t)=[tex]\theta[/tex]0e[tex]\beta[/tex]t
 
  • #7
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.
 
Last edited:

1. How do you determine the magnitude of linear acceleration from angular position equation?

The magnitude of linear acceleration can be determined by using the equation: a = rα, where a is the linear acceleration, r is the radius, and α is the angular acceleration.

2. What is the relationship between linear acceleration and angular position?

The relationship between linear acceleration and angular position is that linear acceleration is directly proportional to the angular position and the radius of the object.

3. Can the magnitude of linear acceleration be negative?

Yes, the magnitude of linear acceleration can be negative. This means that the object is accelerating in the opposite direction of its velocity, which is known as deceleration.

4. How is the magnitude of linear acceleration affected by changes in angular position?

The magnitude of linear acceleration is affected by changes in angular position because it is directly proportional to the angular acceleration. This means that as the object's angular position changes, the magnitude of its linear acceleration also changes.

5. What is the unit of measurement for the magnitude of linear acceleration?

The unit of measurement for the magnitude of linear acceleration is meters per second squared (m/s²). This represents the change in velocity per unit of time and is a measure of how fast the object is accelerating.

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