- #1
mconnor92
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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.