If a rock climber accidentally drops a 52.5-g piton from a height of 325 meters, what would its speed be just before striking th
1 answer:
You might be interested in
Answer:
A) = 1.44 kg m², B) moment of inertia must increase
Explanation:
The moment of inertia is defined by
I = ∫ r² dm
For figures with symmetry it is tabulated, in the case of a cylinder the moment of inertia with respect to a vertical axis is
I = ½ m R²
A very useful theorem is the parallel axis theorem that states that the moment of inertia with respect to another axis parallel to the center of mass is
I = + m D²
Let's apply these equations to our case
The moment of inertia is a scalar quantity, so we can add the moment of inertia of the body and both arms
=
+ 2
= ½ M R²
The total mass is 64 kg, 1/8 corresponds to the arms and the rest to the body
M = 7/8 m total
M = 7/8 64
M = 56 kg
The mass of the arms is
m’= 1/8 m total
m’= 1/8 64
m’= 8 kg
As it has two arms the mass of each arm is half
m = ½ m ’
m = 4 kg
The arms are very thin, we will approximate them as a particle
= M D²
Let's write the equation
= ½ M R² + 2 (m D²)
Let's calculate
= ½ 56 0.20² + 2 4 0.20²
= 1.12 + 0.32
= 1.44 kg m²
b) if you separate the arms from the body, the distance D increases quadratically, so the moment of inertia must increase