Falling Chimney, Angular and Linear Acceleration

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  • #1

Homework Statement



A 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?
R = 49 m
theta = 29°

Homework Equations



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?
 

Answers and Replies

  • #2
29
0
The problem with your solution is that a is not constant when the chimney is falling. You have to use mgh = .5mv^2 to find v and then you can find a tangential. To find the change in height, you must use the center of mass of the chimney ( the middle) when it is straight, and when it it at an angle of 29 degrees)
 
  • #3
alright so I can apply conservation of energy by saying:

1/2 Mgh = 1/2*M*v^2 + 1/2 gh * cos(theta)
gh = v^2 + gh(cos(theta))
v = sqrt( gh(1-cos(theta)) )


But that would give me an answer of 7.76 m/s^2, which is close, but not right (Webassign gives me an answer of 7.13 m/s^2)

*Edit*
1/2 Mgh = 1/2*M*v^2 + 1/2 Mgh * cos(theta)

Forgot the mass on the Potential Energy, but the other two variants of the formula are correct, so this wouldn't account for the error in the final answer.
 
Last edited:

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