193k views
4 votes
A building is being knocked down with a wrecking ball, which is a big metal sphere that swings on a 12-m-long cable. You are (unwisely!) standing directly beneath the point from which the wrecking ball is hung when you notice that the ball has just been released and is swinging directly toward you. How much time do you have to move out of the way?

User Billschen
by
7.9k points

2 Answers

3 votes

Final answer:

An unsupported concrete wall struck near its top with a wrecking ball is likely to fall over by rotating at its base due to greater torque. When still, a wrecking ball has potential energy; if released, it converts this into kinetic energy during its swing.

Step-by-step explanation:

If a wrecking ball hits an unsupported concrete wall near the top, it is much more likely to cause the wall to fall over by rotating at its base rather than falling straight down. This is due to the fact that when an object is struck at a point far from its pivot (in this case, the base of the wall), it generates a larger torque which promotes rotational motion. The leverage effect enables a force applied at a distance from the pivot to be more effective at turning or tipping the object. If the wall is hit at the base with the same force, how it falls will depend on how firmly attached it is at the base. A well-secured wall might still crack and crumble, while a loosely attached wall might tip over more readily.

When considering a wrecking ball that is motionless, it is still associated with energy. This form of energy, called potential energy, is related to the position of the wrecking ball and the force of gravity acting on it. If the ball is released, the potential energy will be transferred into kinetic energy as the wrecking ball accelerates downward and swings. At the highest point of its swing, the ball has maximum potential energy and as it swings down, this energy is converted into kinetic energy, which is at its maximum at the lowest point of the swing.

User SecretIndividual
by
7.3k points
5 votes

Answer:

Time to move out of the way = 1.74 s

Step-by-step explanation:

This case is similar to simple pendulum. The period of simple pendulum is given by
2\pi \sqrt{(l)/(g) }.

Period of wrecking ball
=2\pi \sqrt{(l)/(g)}=2\pi \sqrt{(12)/(9.81)}=6.95s

Time to move out of the way is one fourth of period = 6.95/4 = 1.74 seconds.

Time to move out of the way = 1.74 s

User Healing
by
7.8k points