An object rotates about a fixed axis, and th angular position of a reference line on the object is given by THETA(t)=0.4e^2t, where THETA is in radians, and t is in seconds.
[a.] what is the object's angular acceleration at t = 2 s?
..this is my work so far
THETA(t) = 0.4e^((2)(2))
= 21.8 degrees
21.8 x (Pi/180) = 0.38 rad
should i be using this equation?:
(w2-w1)/(t2-t1) = Angular acceleration
.. and
[b.] What is the tangential AND centripetal acceleration of a point on the object that is 4.0 cm from the axis of rotation?
.. centripetal acceleration's equation would be A = w^2 r
so.. that would be (0.38^2)(4.0) = 0.5776 rad/cm
did i do that right?
and im stuck on how to do the tangential acceleration equation.. can u show me how to work that one out?
thanks for your help!!
You need to take first and second deriviatives of theta(t) to do this problem. Call the angular acceleration "alpha"
theta = 0.4 e^(2t)
omega = d(theta)/dt = 0.8*e^(2t)
alpha = d(omega)/dt = 1.6*e^(2t)
If R is the distance of an object from the axis of rotation,
tangential acceleration = R*(alpha)
centripetal acceleration = R*(omega)^2
To find the object's angular acceleration at t=2s, you have correctly applied the formula for angular acceleration:
(w2 - w1) / (t2 - t1) = Angular acceleration
However, your calculation of theta(t) is incorrect. The correct equation for theta(t) is:
theta(t) = 0.4 * e^(2t)
To find the angular acceleration at t=2s, you need to take the derivatives of theta(t) with respect to t. Let's start by finding the first derivative:
omega(t) = d(theta(t)) / dt
Differentiating theta(t) with respect to t, we have:
omega(t) = 0.4 * d(e^(2t)) / dt
Since the derivative of e^x is e^x, the derivative of e^(2t) with respect to t is 2e^(2t). Thus:
omega(t) = 0.4 * 2e^(2t)
omega(t) = 0.8e^(2t)
Next, we need to find the second derivative to get the angular acceleration:
alpha(t) = d(omega(t)) / dt
Differentiating omega(t) with respect to t, we have:
alpha(t) = 0.8 * d(e^(2t)) / dt
Again, using the derivative of e^x, we get:
alpha(t) = 0.8 * 2e^(2t)
alpha(t) = 1.6e^(2t)
Now we can substitute t=2s into the expression for alpha(t):
alpha(2) = 1.6e^(2*2)
alpha(2) = 1.6e^4
So the object's angular acceleration at t=2s is approximately 1.6e^4.