a force of magnitude 30 n stretches a spring 0.83 m from equilibrium. what is the value of the spring constant?
1 answer:
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Answer:
Δθ₁ = 172.5 rev
Δθ₁h = 43.1 rev
Δθ₂ = 920 rev
Δθ₂h = 690 rev
Explanation:
- Assuming uniform angular acceleration, we can use the following kinematic equation in order to find the total angle rotated during the acceleration process, from rest to its operating speed:
- Now, we need first to find the value of the angular acceleration, that we can get from the following expression:
- Since the machine starts from rest, ω₀ = 0.
- We know the value of ωf₁ (the operating speed) in rev/min.
- Due to the time is expressed in seconds, it is suitable to convert rev/min to rev/sec, as follows:
- Replacing by the givens in (2):
- Solving for α:
- Replacing (5) and Δt in (1), we get:
- in order to get the number of revolutions during the first half of this period, we need just to replace Δt in (6) by Δt/2, as follows:
- In order to get the number of revolutions rotated during the deceleration period, assuming constant deceleration, we can use the following kinematic equation:
- First of all, we need to find the value of the angular acceleration during the second period.
- We can use again (2) replacing by the givens:
- ωf =0 (the machine finally comes to an stop)
- ω₀ = ωf₁ = 57.5 rev/sec
- Δt = 32 s
- Solving for α in (9), we get:
- Now, we can replace the values of ω₀, Δt and α₂ in (8), as follows:
- In order to get finally the number of revolutions rotated during the first half of the second period, we need just to replace 32 s by 16 s, as follows: