Paragraph for question nos. 266 to 268
Have you ever noticed a tower falling down. As it falls down, it often breaks somewhere along its length. The question is why. A safe assumption that can be made is that the tower is hinged at its base. The tangential acceleration of any point on the tower is \( x \alpha \) where \( x \) is distance from hinge and \( \alpha \) the angular acceleration. As it tips further the angular acceleration increases. Eventually, higher portions of tower experience an acceleration greater than that of gravity. For this high acceleration there should be a force in addition to gratitational force. This comes from shear force by lower portions of tower. At the point when shear forces cannot provide for such acceleration the tower breaks.
The tower can be modelled to be a uniform rod of length \( L \) and mass \( m \) for the questions below.
The angular acceleration of the tower increases as it falls because.
(A) the moment of inertia decreases
(B) the force of gravity increases
(C) the lever arm of gravitational force increases.
(D) the lever arm of gravitational force decreases.
267. At what angle from the vertical does the tangential acceleration of the top of tower becomes \( 1.2 g \) ?
(A) \( 37^{\circ} \)
(B) \( 53^{\circ} \)
(C) \( 60^{\circ} \)
(D) \( 30^{\circ} \)
268. If the tower breaks at \( L / 3 \) from top when angle made by tower with vertical is \( 60^{\circ} \), what is shear stress at the point of breaking?
(A) \( \frac{11 mg }{8 \sqrt{3 A }} \)
(B) \( \frac{m g}{2 \sqrt{3} A} \)
(C) \( \frac{m g}{8 \sqrt{3} A } \)
(D) none of these
