Question

* A tall, cylindrical chimney falls over when its base is ruptured. Treat the chimney as a thin rod of length 63.0 m. Answer the following for the instant it makes an angle of θ with the vertical as it falls.

1. Calculate the angular speed at the angle θ =50.0°.

2. Calculate the tangential speed at the angle θ =50.0° of the top.

3. Calculate the tangential speed at the angle θ =50.0° at the center of mass of the chimney.

4. Calculate the angular acceleration at the angle θ =50.0°.

5. Calculate the tangential acceleration at the angle θ =50.0° of the top.

6. Calculate the radial acceleration at the angle θ =50.0° of the top

Answer 1

Apply conservation of energy

Ei = Ef

mgh_com = mgh' + 1/2 I w^2

mgH/2 = mg H/2 cos theta + 1/2 ( 1/3 mH^2) w^2

w= sqrt 3g( 1- cos theta)/H

= sqrt 3(9.8) (1-cos50)/63

=0.40 rad/s

(b)

v= rw = Lw = 63 ( 0.40) = 25.72 m/s

(c) same as in part because tangential speed does not change

(d)

alpha = dw/dt

= d/dt ( sqrt 3g( 1- cos theta)/H

= ( 1/2 ( 1/sqrt 3g( 1- cos theta)/H) 3g/H sin theta) d theta/dt

= 1/2 (1/sqrt 3g( 1- cos theta)/H) 3g/H sin ttheta) w

alpha = 3g/2H sin tehta

= 3(9.8)/ 2( 60.3) * sin 50

=0.186 rad/s^2

(5)

at= r alpha = 60.3 (0.186) = 11.26 m/s^2

(6)

ar= rw^2 = 60.3) ( 0.40)^2 =9.648 m/s^2