Using the formula for kinetic energy of a moving particle k=12mv2, find the kinetic energy ka of particle a and the kinetic energy kb of particle
b. remember that both particles rotate about the y axis. express your answers in terms of m, ω, and r separated by a comma.
Answers
Complete Question
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Answer:
The kinetic energy for particle a is [tex]K_a = \frac{9}{2} mr^2 w^2[/tex]
The kinetic energy for particle b is [tex]K_b = m w^2r^2 [/tex]
Explanation:
From the question we are told that
The mass of particle a is [tex]m[/tex]
The mass of particle b is [tex]2 m[/tex]
Generally the kinetic energy for particle a is mathematically represented as
[tex]K_a = \frac{1}{2} m v^2[/tex]
Here [tex]v_a[/tex] is the linear velocity of particle a which is mathematically represented as
[tex]v_b = w * 3r[/tex]
So
[tex]K = \frac{1}{2} m {w * 3r}^2[/tex]
=> [tex]K_a = \frac{9}{2} mr^2 w^2[/tex]
Generally the kinetic energy for particle a is mathematically represented as
[tex]K_b = \frac{1}{2} 2m v_b^2[/tex]
Here [tex]v_b[/tex] is the linear velocity of particle a which is mathematically represented as
[tex]v_b = w * r[/tex]
So
[tex]K_b = m (wr)^2[/tex]
=> [tex]K_b = m w^2r^2[/tex]
The kinetic energy of the particle in terms of angular speed (ω), mass of the object (m) and the radius of the path is [tex]\frac{1}{2} m \omega ^2 r^2[/tex].
The given parameters;
kinetic energy of the particle, K.E = ¹/₂mv²The kinetic energy of the of the particle in terms of the angular velocity is calculated as follows;
[tex]K_b = \frac{1}{2} m (\omega r)^2\\\\K_b = \frac{1}{2} m \omega ^2 r^2[/tex]
where;
m is the mass of the objectω is the angular speed of the objectr is the radius of the circular pathThus, the kinetic energy of the particle in terms of angular speed (ω), mass of the object (m) and the radius of the path is [tex]\frac{1}{2} m \omega ^2 r^2[/tex].
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Answer:
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Answer:
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Answer:
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Answer:
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Answers
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Answer:
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Answer:
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Answer:
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Answers
Answer:
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Answers
Answer:
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Answer:
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Answer:
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Answer:
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Answers
Answer:
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Answers
Answer:
Driveway across the street ([tex]\theta = 6^{\circ}[/tex]) is the only choice for a safe parking.
Explanation:
Let suppose that car is represented by a particle, then we proceed to construct its free body diagram and corresponding equations of equilibrium:
[tex]\Sigma F_{x'} = f - W\cdot \sin \theta = 0[/tex] (Eq. 1)
[tex]\Sigma F_{y'} = N-W\cdot \cos \theta = 0[/tex] (Eq. 2)
Where:
[tex]f[/tex] - Static friction force, measured in newtons.
[tex]N[/tex] - Normal force on the car from the ground, measured in newtons.
[tex]W[/tex] - Weight of the car, measured in newtons.
[tex]\theta[/tex] - Driveway inclination, measured in sexagesimal degrees.
By applying definitions of maximum static friction force and weight, we expand the system of equations presented above:
[tex]\mu_{s}\cdot N -m\cdot g\cdot \sin \theta = 0[/tex] (Eq. 1b)
[tex]N - m\cdot g \cdot \cos \theta = 0[/tex] (Eq. 2b)
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[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
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[tex]\mu_{s}\cdot m\cdot g \cdot \cos \theta -m\cdot g \cdot \sin \theta = 0[/tex]
[tex]\mu_{s} = \tan \theta[/tex]
[tex]\theta = \tan^{-1}\mu_{s}[/tex] (Eq. 3)
Based on such result, we can conclude that if driveway inclination is greater than value reported, then car shall not be safe in case of parking. Our reference angle is: ([tex]\mu_{s} = 0.15[/tex])
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By direct comparison, we find that driveway across the street ([tex]\theta = 6^{\circ}[/tex]) is the only choice for a safe parking.
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Answers
Answer:
Unbalanced forces are not examples of Newton's third law because not all opposite reactions are unbalanced forces.
Explanation: