A tall, cylindrical chimney falls over when its base is ruptured. Treat the chimney as a thin rod of length 58.8 m. At the instant it makes an angle of 30.2° with the vertical as it falls, what are (a) the radial acceleration of the top, and (b) the tangential acceleration of the top. (Hint: Use energy considerations, not a torque.) (c) At what angle θ is the tangential acceleration equal to g? Assume free-fall acceleration to be equal to 9.81 m/s2.

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Xema8 Xema8 · Answer 1 · 2020-05-10T07:26:26+02:00

Answer:

Explanation:

Using conservation of mechanical energy for rotational motion of chimney

gain of rotational KE = loss of potential energy

1/2 Iω² = mgl/2 ( 1 - cosθ ) ; I is moment of inertia , ω is angular velocity , m is mass of chilney , l is its length .

I = 1/3 m l²

1/6 m l²ω²= mgl/2 ( 1 - cosθ )

1/6 m v² = mgl/2 ( 1 - cosθ )

v² = 3gl( 1 - cosθ )

radial acceleration at the top = v² / r

=3gl( 1 - cosθ ) / l

= 3g ( 1 - cosθ )

= 3 x 9.8 ( 1 - cos 30.2 )

= 4 m / s²

b ) Let angular acceleration be α .

Torque acting at that time

mgsinθ x l /2 = I α

mgsinθ x l /2 = 1/3 m l² α

α = 3g sinθ / 2l

tangential acceleration

α x l = 3gsinθ / 2

= 3 x 9.8 sin 30.2 / 2

= 7.4 m /s²

c ) tangential acceleration equal to g

3gsinθ / 2 = g

sinθ = 2 / 3

42 degree .