Question

A rod with a length of L = 1 m is held vertically so that one end rests on the floor. After that

let go of the stick and fall. Let us assume that the supported end does not slip and that the thickness of the bar is very small compared to its length (that is, its moment of inertia with respect to the center of gravity is IT = 1/12mL^2

(a) With what speed does the other end of y hit the floor?
(b) How much does the end impact speed increase if it were a 100 m tall object that was originally
blasted off (and not broken when falling)?
Hint: Use the law of conservation of mechanical energy. Kinetic energy can be written in two ways:
1. Pure rotary movement around the end of the rod. 2. Rotary movement around the center of gravity + movement of the center of gravity. Both procedures,
of course, give the same result (you can try both and see for yourself).

Answer 1

Step-by-step explanation:

(a) Let's use the law of conservation of mechanical energy to determine the speed with which the other end of the rod hits the floor. When the rod is released, it begins to rotate around its center of gravity and falls to the floor. At the moment of release, the rod has no kinetic energy or potential energy, but it has potential energy when it reaches the floor. The energy is conserved, so we can equate the initial potential energy to the final kinetic energy.

The initial potential energy of the rod is given by:

U_i = mgh

where m is the mass of the rod, g is the acceleration due to gravity, and h is the height of the center of gravity above the floor. Since the rod is vertical, h = L/2. The mass of the rod can be calculated using its density ρ and cross-sectional area A:

m = ρAL

The final kinetic energy of the rod is given by:

K_f = (1/2)Iω^2 + (1/2)mv^2

where I is the moment of inertia of the rod with respect to its center of gravity, ω is the angular velocity of the rod, and v is the linear velocity of the center of gravity. At the moment when the rod hits the floor, the angular velocity is zero, so the first term in the above equation is zero. We can simplify the equation to:

K_f = (1/2)mv^2

We can equate the initial potential energy and final kinetic energy to get:

mgh = (1/2)mv^2

Solving for v, we get:

v = sqrt(2gh)

Substituting the given values, we get:

v = sqrt(2gL/2) = sqrt(gL/2)

Now, we can substitute the values of g and L to get:

v = sqrt(9.81 m/s^2 x 1 m/2) = sqrt(4.905) m/s

Therefore, the other end of the rod hits the floor with a speed of approximately 2.216 m/s.

(b) If the length of the rod were 100 m instead of 1 m, the speed with which the end hits the floor would increase significantly. The potential energy of the rod when it is released is proportional to its height above the floor, so when the length of the rod is increased by a factor of 100, the potential energy increases by a factor of 100 as well. Therefore, the final speed of the end hitting the floor would be:

v' = sqrt(2gh') = sqrt(2g(100L)/2) = sqrt(100gL/2) = 10sqrt(gL/2)

Substituting the given values, we get:

v' = 10sqrt(9.81 m/s^2 x 100 m/2) = 10sqrt(490.5) m/s

Therefore, the end of the 100 m tall object would hit the floor with a speed of approximately 70.0 m/s, which is a significant increase compared to the initial speed of the 1 m rod.