5. A belt runs on a wheel of 30.0-cm radius. During the time that the wheel coasts uniformly to rest from an initial speed of 2.
00 rev/s, 25.0 m of belt length passes over the wheel. Find the angular acceleration of the wheel and the total number of revolutions it turns before stopping.
1 answer:
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Answer:
The speed of the ambulance is 4.30 m/s
Explanation:
Given:
Frequency of the ambulance, f = 1790 Hz
Frequency at the cyclist, f' = 1780 Hz
Speed of the cyclist, v₀ = 2.36 m/s
let the velocity of the ambulance be 'vₓ'
Now,
the Doppler effect is given as:
where, v is the speed of sound
since the ambulance is moving towards the cyclist. thus, the sign will be positive
thus,
on substituting the values, we get
or
vₓ = 4.30 m/s
Hence, <u>the speed of the ambulance is 4.30 m/s</u>
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: