4 identical coins of mass M and radius R are placed in a square, so the center of each coin lies on a corner of the square. The
coins are touching and stuck to each other. What is the moment of inertia of the system about an axis through the center of the square? Consider the parallel axis theorem, and treat coins as disks of uniform mass.
2 answers:
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:
The moment of inertia in point A is:
The, for the 4 coins the moment of inertia is:
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
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Answer:
Speed of the spacecraft right before the collision: .
Assumption: the earth is exactly spherical with a uniform density.
Explanation:
This question could be solved using the conservation of energy.
The mechanical energy of this spacecraft is the sum of:
- the kinetic energy of this spacecraft, and
- the (gravitational) potential energy of this spacecraft.
Let denote the mass of this spacecraft. At a distance of
from the center of the earth (with mass
), the gravitational potential energy (
) of this spacecraft would be:
.
Initially, (the denominator of this fraction) is infinitely large. Therefore, the initial value of
will be infinitely close to zero.
On the other hand, the question states that the initial kinetic energy () of this spacecraft is also zero. Therefore, the initial mechanical energy of this spacecraft would be zero.
Right before the collision, the spacecraft would be very close to the surface of the earth. The distance between the spacecraft and the center of the earth would be approximately equal to
, the radius of the earth.
The of the spacecraft at that moment would be:
.
Subtract this value from zero to find the loss in the of this spacecraft:
Assume that gravitational pull is the only force on the spacecraft. The size of the loss in the of this spacecraft would be equal to the size of the gain in its
.
Therefore, right before collision, the of this spacecraft would be:
.
On the other hand, let denote the speed of this spacecraft. The following equation that relates
and
to
:
.
Rearrange this equation to find an equation for :
.
It is already found that right before the collision, . Make use of this equation to find
at that moment:
.