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serg [7]
3 years ago
5

The rotational kinetic energy term is often called the kinetic energy in the center of mass, while the translational kinetic ene

rgy term is called the kinetic energy of the center of mass. You found that the total kinetic energy is the sum of the kinetic energy in the center of mass plus the kinetic energy of the center of mass. A similar decomposition exists for angular and linear momentum. There are also related decompositions that work for systems of masses, not just rigid bodies like a dumbbell. It is important to understand the applicability of the formula Ktot=Kr+Kt. Which of the following conditions are necessary for the formula to be valid? Check all that apply.
a. The velocity vector V must be perpendicular to the axis of rotation.
b.The velocity vector must be perpendicular or parallel to the axis of rotation.
c.The moment of inertia must be taken about an axis through the center of mass.
Physics
2 answers:
weqwewe [10]3 years ago
6 0

Answer:

C

Explanation:

The total kinetic energy is the sum of the kinetic energy in the center of mass (Rotational Kinetic energy) plus the kinetic energy of the center of mass( Translational Kinetic Energy).

The formula

K_{tot} = K_{t} +K_{r}  is applicable only when

The moment of inertia must be taken about an axis through the center of mass.

Aneli [31]3 years ago
5 0

Answer:

c.The moment of inertia must be taken about an axis through the center of mass.

Explanation:

The movements of rigid bodies can always be divided into the translation movement of the center of mass and that of rotation around the center of mass. However, we can demonstrate that this is true for the kinetic energy of a rigid body that has both translational and rotational movement.

In this case the kinetic energy of the body is the sum of a part associated with the movement of the center of mass and another part associated with the rotation about an axis passing through the center of mass. This is all represented by the form Ktot = Kr + Kt, however we must consider that the moment of inertia must be taken around an axis through the center of mass, since rigid bodies at rest tend to remain at rest.

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A bike with tires of radius 0.330 m speeds up from rest to 5.33 m/s in 6.27 s. What is the angular acceleration of the tires?
Nimfa-mama [501]

The angular acceleration of a bike with tires of radius 0.330 m speeds up from rest to 5.33 m/s in 6.27 s will be 2. 57 rad/s²

<h3>What is angular acceleration?</h3>

Angular acceleration can be defined as the time it takes for a change in angular velocity. It is denoted as 'α' with a unit of rads/s²

It is expressed thus;

α= Δω ÷ Δt

Where α = angular acceleration

Δw = change in angular velocity = velocity ÷ radius

Δ t = change in time

<h3>How to calculate the angular acceleration</h3>

Using the formula:

α= Δω ÷ Δt

v = 5. 33m/s, r = 0. 33m and t = 6.27s

Substitute the values to get the angular velocity

Δw = v÷ r = 5. 33 ÷ 0.330 = 16. 15 m/s

Substitute the value of Δw into the equation

α= Δω ÷ Δt = 16. 15 ÷ 6. 27 = 2. 57 rad/s²

Therefore, the angular acceleration of a bike with tires radius of 0. 330m, speed of 5. 33mls in 6. 27s is 2. 57 rad/s²

Learn more about angular acceleration here:

brainly.com/question/21278452

#SPJ1

8 0
1 year ago
Detailed measurements of the disk and central bulge region of our Galaxy suggest our Milky Way is a:Select one:A. quasar. B. bar
chubhunter [2.5K]

I believe the answer is:

~D. Very dusty irregular galaxy.

Hope this helps!!!

4 0
2 years ago
5. A massless string passes over a frictionless pulley and carries
devlian [24]

Answer:

2m₁m₃g / (m₁ + m₂ + m₃)

Explanation:

I assume the figure is the one included in my answer.

Draw a free body diagram for each mass.

m₁ has a force T₁ up and m₁g down.

m₂ has a force T₁ up, T₂ down, and m₂g down.

m₃ has a force T₂ up and m₃g down.

Assume that m₁ accelerates up and m₂ and m₃ accelerate down.

Sum of the forces on m₁:

∑F = ma

T₁ − m₁g = m₁a

T₁ = m₁g + m₁a

Sum of the forces on m₂:

∑F = ma

T₁ − T₂ − m₂g = m₂(-a)

T₁ − T₂ − m₂g = -m₂a

(m₁g + m₁a) − T₂ − m₂g = -m₂a

m₁g + m₁a + m₂a − m₂g = T₂

(m₁ − m₂)g + (m₁ + m₂)a = T₂

Sum of the forces on m₃:

∑F = ma

T₂ − m₃g = m₃(-a)

T₂ − m₃g = -m₃a

a = g − (T₂ / m₃)

Substitute:

(m₁ − m₂)g + (m₁ + m₂) (g − (T₂ / m₃)) = T₂

(m₁ − m₂)g + (m₁ + m₂)g − ((m₁ + m₂) / m₃) T₂ = T₂

(m₁ − m₂)g + (m₁ + m₂)g = ((m₁ + m₂ + m₃) / m₃) T₂

m₁g − m₂g + m₁g + m₂g = ((m₁ + m₂ + m₃) / m₃) T₂

2m₁g = ((m₁ + m₂ + m₃) / m₃) T₂

T₂ = 2m₁m₃g / (m₁ + m₂ + m₃)

8 0
3 years ago
If I have an object that starts from rest and reaches a velocity of 25m/s over a distance of 11m what is the acceleration of the
Elodia [21]

Answer:

Here is something that may help you!!

Explanation:

I found it in a cite (not that I'm plagiarizing, or anything).

3 0
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Match the choices to the appropriate blank . Use each choice only once.
pav-90 [236]

Answer:

1. about 1.5 AU

2. about 5 AU

3. about 8 light-years

4. about 100,000 light-years

5. less than 0.01 AU

Explanation:

a. Mars is about 1.5 AU from the Sun.

b. Jupiter is about 5 AU from the Sun.

c. The star Sirius is about 8 light-years from the Sun.

d. The diameter of the Milky Way Galaxy is about 100,000 light-years.

e. The distance from Earth to the Moon is less than 0.01 AU.

Note: AU is an acronym for Astronomical Unit and it is a standard unit by astronomers to illustrate the distance between the planetary bodies found in the solar system.

5 0
2 years ago
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