Question

A 2.10-m-long pole is balanced vertically on its tip. It starts to fall and its lower end does not slip. Part A What will be the speed of the upper end of the pole just before it hits the ground?

Answer 1

The speed of the upper end of the pole just before it hits the ground is 6.48 m/s.

Step-by-step explanation:

The speed of the upper end of the pole just before it hits the ground can be determined by considering the conservation of energy. As the pole falls, its potential energy is converted into kinetic energy. At the top of its fall, all of the potential energy is converted into kinetic energy, so we can equate the two:

mg(2h) = 1/2mv^2

where m is the mass of the pole, g is the acceleration due to gravity, h is the height of the pole, and v is the speed of the upper end of the pole. Rearranging this equation, we get:

v = sqrt(2gh)

Substituting the given values, we find:

v = sqrt(2 * 9.8 * 2.10)

v = 6.48 m/s

Answer 2

The speed of the upper end of the pole just before it hits the ground is 4.67 m/s.

Step-by-step explanation:

In this scenario, we can use the principles of physics to find the speed of the upper end of the pole just before it hits the ground. When the pole is released, it will start to rotate around its lower end as it falls. Since the lower end does not slip, the linear velocity of the upper end is equal to the angular velocity of the pole multiplied by its length.

We can calculate the angular velocity of the pole using the equation:

angular velocity = (acceleration due to gravity) / (length of the pole)

Given that the length of the pole is 2.10 m, the acceleration due to gravity is 9.8 m/s², we can substitute these values into the equation:

angular velocity = 9.8 m/s² / 2.10 m = 4.67 rad/s

Therefore, the linear velocity of the upper end of the pole just before it hits the ground is 4.67 m/s.