A particle of mass m moves under an attractive central force F(r) = -Kr4 with angular momentum L. For what energy will the motio
n be circular, and what is the radius of the circle? Find the frequency of radial oscillations in the particle is given a small radial impulse.
1 answer:
Answer:
Angular velocity is same as frequency of oscillation in this case.
ω = x
Explanation:
- write the equation F(r) = -K with angular momentum <em>L</em>
- Get the necessary centripetal acceleration with radius r₀ and make r₀ the subject.
- Write the energy of the orbit in relative to r = 0, and solve for "E".
- Find the second derivative of effective potential to calculate the frequency of small radial oscillations. This is the effective spring constant.
- Solve for effective potential
- ω = x
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Answer: The end point of a spring oscillates with a period of 2.0 s when a block with mass m is attached to it. When this mass is increased by 2.0 kg, the period is found to be 3.0 s. Then the mass m is 0.625kg.
Explanation: To find the answer, we need to know more about the simple harmonic motion.
<h3>What is simple harmonic motion?</h3>- A particle is said to execute SHM, if it moves to and fro about the mean position under the action of restoring force.
- We have the equation of time period of a SHM as,
- Where, m is the mass of the body and k is the spring constant.
- Given that,
- We have to find the value of m,
Thus, we can conclude that, the mass m will be 0.625kg.
Learn more about simple harmonic motion here:
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1) He has both potential and kinetic energy
2) Before the parachute opens, the potential energy decreases and the kinetic energy increases
Explanation:
1)
The gravitational potential energy of a body is the energy possessed by the object due to its position in a gravitational field, and it is given by:
where
m is the mass of the body
g is the acceleration of gravity
h is the height of the body above the ground
On the other hand, the kinetic energy of a body is the energy possessed by the body due to its motion; it is given by
where
v is the speed of the object
Here Mr. Leppold has both potential and kinetic energy before opening the parachute, because:
- It is moving at a certain speed, so , therefore he has kinetic energy
- He is at a certain height above the ground, , therefore he has potential energy
2)
The total mechanical energy of Mr.Leppold is the sum of the potential and the kinetic energy:
According to the law of conservation of energy, in absence of air resistance, this quantity remains constant.
During the fall, the height of Leppold decreases: this means that as decreases, the potential energy decreases too.
However, the total energy E must remain constant: therefore, this means that the kinetic energy KE must increase, and this occurs because the speed of Mr. Leppold increases as he falls.
Learn more about kinetic and potential energy:
#LearnwithBrainly