A planet has a period of revolution about the Sun equal to and a mean distance from the Sun equal to R .T^2 varies directly as _
1 answer:
You might be interested in
Horizontal speed = 24.0 m/s
height of the cliff = 51.0 m
For the initial vertical speed will are considering the vertical component. Therefore,
Since the student fires the canonical ball at the maximum height of 51 m, the initial vertical velocity will be zero. This means
let's find how long the ball remained in the air.
Finally, let's find the how far from the base of the building the ball landed(horizontal distance)
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 = + 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
= 4 I
I_{total} = 4 5/2 M R²
I_{total} = 10 M R²