What is the moment of inertia of a cube with mass M=0.500kg and side lengths s=0.030m about an axis which is both normal (perpen
dicular) to and through the center of a face of the cube? Watch your units. *Note that moment of inertia is a purely geometric property of a rigid body, and in lab we will call this the static moment of inertia. We will also experimentally determine a dynamic moment of inertia which will be measured from the rigid body's angular acceleration.
1 answer:
Answer:
The moment of inertia I is
I = 2.205x10^-4 kg/m^2
Explanation:
Given mass m = 0.5 kg
And side lenght = 0.03 m
Moment of inertia I = mass x radius of rotation squared
I = mr^2
In this case, the radius of rotation is about an axis which is both normal (perpendicular) to and through the center of a face of the cube.
Calculating from the dimensions of the the box as shown in the image below, the radius of rotation r = 0.021 m
Therefore,
I = 0.5 x 0.021^2 = 2.205x10^-4 kg/m^2
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Answer:
the ratio of Hank's mass to Harry's mass is 0.7937 or [ 0.7937 : 1
Explanation:
Given the data in the question;
Hank and Harry are two ice skaters, since both are on top of ice, we assume that friction is negligible.
We know that from Newton's Second Law;
Force = mass × Acceleration
F = ma
Since they hold on to opposite ends of the same rope. They have the same magnitude of force |F|, which is the same as the tension in the rope.
Now,
Mass × Acceleration
= Mass
× Acceleration
so
Mass / Mass
= Acceleration
/ Acceleration
given that; magnitude of Hank's acceleration is 1.26 times greater than the magnitude of Harry's acceleration,
Mass / Mass
= 1 / 1.26
Mass / Mass
= 0.7937 or [ 0.7937 : 1 ]
Therefore, the ratio of Hank's mass to Harry's mass is 0.7937 or [ 0.7937 : 1 ]
Answer:
15.32°
Explanation:
We have given the wavelength
Diffraction grating is 1460 lines per cm
So (as 1 m=100 cm )
For maximum diffraction
here m is order of diffraction
So
Answer:
a) E_total = 6,525 10⁴ N /C ,field direction is radial outgoing
b) E_total = 1.89 10⁶ N / C, field is incoming radial
c) E_total = 0
Explanetion:
For this exercise we can use that the charge in a spherical shell can be considered concentrated at its center and that the electric field inside the shell is zero, since Gauss's law is
Ф = E .dA = q_{int} /ε₀
inside the spherical shell there are no charges
The electric field is a vector quantity, so we calculate the field created by each shell and add it vectorly.
We have two sphere shells with radii 0.050m and 0.15m respectively
a) point where you want to know the electric field d = 0.20 m
shell 1
the point is on the outside,d>ro, therefore we can consider the charge to be concentrated in the center
E₁ = k q₁ / d²
shell 2
the point is on the outside,d>ro
E₂ = k q₂ / d²
the total camp is
E_total = -E₁ + E₂
E_total = k ( )
E_total = 9 10⁹ (-2.1 10⁻⁶+ 5 10⁻⁶ / .2²
E_total = 6,525 10⁵ N /C
The field direction is radial and outgoing ti the shells
b) the calculation point is d = 0.10m
shell 1
point outside the shell d> ro
E₁ = k q₁ / d²
shell 2
the point is inside the shell d <ro
Therefore, according to Gauss's law, since there are no charges in the interior, the electrioc field is zero
E₂ = 0
E_total = E₁
E_total = k q₁ / d²
E_total = 9 10⁹ 2.1 10⁻⁶ / 0.1²
E_total = 1.89 10⁶ N / A
As the charge is negative, this field is incoming radial, that is, it is directed towards the shell 1
c) the point of interest d = 0.025 m
shell 1
point is inside the shell d< ro
as there are no charges inside
E₁ = 0
shell 2
point is inside the radius of the shell d <ro
E₂ = 0
the total field is
E_total = 0