Respuesta :
Answer:
I_{total} = 10 M R²
Explanation:
The concept of moment of inertia in rotational motion is equivalent to the concept of inertial mass for linear motion. The moment of inertia is defined
I = ∫ r² dm
For body with high symmetry it is tabulated, in these we can simulate them by a solid disk, with moment of inertia for an axis that stops at its center
I = ½ M R²
As you hear they ask for the moment of energy with respect to an axis parallel to the axis of the disk, we can use the theorem of parallel axes
I = [tex]I_{cm}[/tex] + M D²
Where I_{cm} is the moment of inertia of the disk, M is the total mass of the system and D is the distance from the center of mass to the new axis
Let's apply these considerations to our problem
The moment of inertia of the four discs is
I_{cm} = I
I_{cm} = ½ M R²
For distance D, let's use the Pythagorean Theorem. As they indicate that the coins are touched the length of the square is L = 2R, the distance from any spine to the center of the block is
D² = (R² + R²)
D² = R² 2
Let's calculate the moment of inertia of a disk with respect to the axis that passes through the center of the square
I = ½ M R2 + M R² 2
I = 5/2 M R²
This is the moment of inertia of a disc as we have four discs and the moment of inertia is a scalar is additive, so
[tex]I_{total}[/tex] = 4 I
I_{total} = 4 5/2 M R²
I_{total} = 10 M R²
Answer:
The moment of inertia for the 4 coins is 10*MR²
Explanation:
According the attached diagram, the moment of inertia in point O is:
[tex]I_{O} =\frac{1}{2} MR^{2}[/tex]
The moment of inertia in point A is:
[tex]I_{A} =\frac{1}{2} MR^{2} +M(OA)^{2} \\OA=\frac{1}{2} \sqrt{2} BO=\frac{1}{2} \sqrt{2}(2R)=\sqrt{2} R\\I_{A}=\frac{1}{2} MR^{2}+2MR^{2} =\frac{5}{2} MR^{2}[/tex]
The, for the 4 coins the moment of inertia is:
[tex]I_{4c} =4*I_{A} =4*\frac{5}{2} MR^{2} =10MR^{2}[/tex]
