R = 49 mA tall, cylindrical chimney falls over when its base is ruptured. Treat the chimney as a thin rod of length 49.0 m. Answer the following for the instant it makes an angle of 29.0° with the vertical as it falls. (Hint: Use energy considerations, not a torque.)
(a) What is the radial acceleration of the top?
(b) What is the tangential acceleration of the top?
(c) At what angle θ is the tangential acceleration equal to g?
theta = 29°
Kinematic equations for Rotational Forces and Circular motion
Kinetic Energy = 1/2*I*w^2
The Attempt at a Solution
First I can find the moment of Inertia for a rod, which is: 1/12 * MR^2
However I need to use the parallel axis theorem to find: 1/12*MR^2 + M(R/2)^2
Simplifying this I get:
I = 1/3 * M * R^2
I already know that I can find the angular velocity through conservation of energy:
KE + U initial = KE + U final
(these can be found if we follow the kinematics of the center of mass)
0 + Mg(h/2) = 1/2*I*w^2 + Mg(R/2 * cos(theta))
g = 1/3 * R * w^2 + g*cos(theta
g*(1-cos(theta)) = 1/3 * R * w^2
w^2 = (3g * (1-cos(theta))) / R
From circular motion kinematics we know that:
a(radial) = (v^2) / R
We also know that:
v = w*R
So combining them I get:
a(radial) = (w*r)^2 / R
a(radial) = w^2 * R
After plugging in values I get:
a(radial) = 3 * g * (1 - cos(theta))
a(radial) = 3 * 9.81 * (1 - cos(29))
a(radial) = 3.69 m/s^2
From there I need to find the tangential acceleration, but I can't seem to get a decent answer.
I know that:
a(tangential) = alpha * R
and using kinematic equations I can find that
w^2 = 2(alpha)*(theta)
alpha = (w^2) / (2*(theta'))
Where theta' here would be a radian measure or 29*(pi/180)
but this ends up giving me a value of 0.0744, which can't be right. Am I missing something?