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Question:
The problem shows a vertical-section U-shaped tube with columns K, L, and M filled with water to different heights h_k, h_l, and h_m. The tube is rotated around the horizontal axis xx' at a constant angular velocity. The points K, L, M have distance d and 2d as given in the figure. We are asked to determine the relationship between the heights h_k, h_l, and h_m.
Answer:
Problem Summary
- The U-shaped tube has three vertical columns at points K, L, and M.
- The tube rotates around the xx' axis with a constant angular velocity.
- The heights of the water columns at K, L, and M are h_k, h_l, and h_m respectively.
- Positions relative to the axis: K is at distance d, L is at zero, and M is at 2d from the rotation axis.
- The water surface levels become different due to the rotation.
Concepts Involved
When a fluid rotates at an angular velocity \omega, the free surface of the fluid forms a paraboloid of revolution because of the balance between pressure and centrifugal force.
The height of the fluid surface at a radial distance r from the axis of rotation is given by the equation:
h(r) = h_0 + \frac{\omega^2 r^2}{2g}
Where:
- h(r) is the height of the fluid at radius r,
- h_0 is the reference height at r=0,
- \omega is the angular velocity,
- g is acceleration due to gravity.
Note: The surface curves upwards away from the axis because the outward centrifugal acceleration increases fluid height.
Step-by-Step Solution
Given:
- r_k = d at K,
- r_l = 0 at L,
- r_m = 2d at M.
Using the formula for the heights:
- Height at K:
h_k = h_0 + \frac{\omega^2 d^2}{2g}
- Height at L (on axis):
h_l = h_0 + \frac{\omega^2 \times 0^2}{2g} = h_0
- Height at M:
h_m = h_0 + \frac{\omega^2 (2d)^2}{2g} = h_0 + \frac{\omega^2 4 d^2}{2g} = h_0 + 2 \frac{\omega^2 d^2}{g}
Breaking down relations:
- h_l is the smallest height because it is on the axis with r=0.
- h_m is the largest since it is at 2d and the height increases with r^2.
- h_k is between h_l and h_m because d < 2d.
Therefore, the relationship is:
h_l < h_k < h_m
or
h_l < h_k < h_m
Answer choice matching:
- A) h_k = h_l = h_m — Incorrect, heights differ due to rotation.
- B) h_l > h_k = h_m — Incorrect.
- C) h_k > h_m > h_l — Incorrect, h_m > h_k.
- D) h_l > h_m > h_k — Incorrect.
- E) h_l < h_k < h_m — Correct.
Summary Table
| Point | Distance from axis (r) | Height formula | Relationship |
|---|---|---|---|
| L | 0 | h_l = h_0 | smallest height |
| K | d | h_k = h_0 + \frac{\omega^2 d^2}{2g} | intermediate height |
| M | 2d | h_m = h_0 + 2 \frac{\omega^2 d^2}{g} | largest height |
Final conclusion:
The correct relationship between water heights in columns K, L, M after rotation is:
h_l < h_k < h_m
Corresponding to choice E) in the question.
If you want, I can also provide the detailed mathematical derivation or help with similar problems! @Mr.Yasin
Çözüm:
Dönen sıvılar incelendiğinde serbest yüzeyin dönüş ekseninden uzaklığa bağlı olarak parabolik bir biçim aldığı bilinmektedir. Dönme kartezyeninde serbest yüzey yüksekliği
z(r)=z_0+\frac{\omega^2\,r^2}{2g}
formülü ile verilir. Burada
- r : Dönme ekseninden uzaklık,
- \omega : Açısal hız,
- g : Yerçekimi ivmesi,
- z_0 : Eksendeki referans yükseklik.
Şekilde K, L, M kollarının dönme eksenine olan uzaklıkları sırasıyla
r_K<r_L<r_M
olduğuna göre serbest yüzey yükseklikleri de şu sırayı izler:
z_K<z_L<z_M
Bağlı kapta kesit alanları eşit olduğundan sıvı yükseklikleri de aynı ilişkiyi taşır:
h_K < h_L < h_M
Buna göre doğru seçenek: E) h_M > h_L > h_K.
Ayt Fizik formül ve konuşarına göre çözer misin?
Soru: Resimli soruda verilen dönmekte olan kapta su yükseklikleri h_k, h_l, h_m arasındaki ilişki nedir?
Detaylı Çözüm ve Açıklama
1. Sorunun Özeti
- Üç kollu, dikey olarak açılmış bir U-boru var: K, L ve M noktalarında.
- Boru, xx' ekseni etrafında sabit açısal hızla dönüyor.
- K, L, M noktalarının dönme eksenine olan uzaklıkları sırasıyla d, 0, ve 2d olarak verilmiş.
- Sıvı seviyeleri h_k, h_l, h_m şeklinde ölçülmüş.
- Amaç: h_k, h_l, h_m arasındaki büyüklük ilişkisini belirlemek.
2. Fiziksel Temel: Dönmekte Olan Sıvının Serbest Yüzeyi
Dönen bir sıvıda, serbest yüzey merkezden uzaklaştıkça parabolik şekilde yükselir. Bunun sebebi merkezkaç kuvvetidir.
Dönme ekseninden r uzaklıktaki serbest yüzey yüksekliği:
h(r) = h_0 + \frac{\omega^2 r^2}{2g}
Burada:
- h_0 = r=0 noktasındaki sıvı yüksekliği (referans),
- \omega = açısal hız (rad/s),
- g = yerçekimi ivmesi (\approx 9.8 \, m/s^2),
- r = dönme ekseninden uzaklık.
Yani yükseklik r^2 ile orantılı olarak artar.
3. Problemin Bilgileri ve Matematği
- K noktası: r_k = d
- L noktası: r_l = 0
- M noktası: r_m = 2d
Formüllere göre:
- $
h_l = h_0 + \frac{\omega^2 \cdot 0^2}{2g} = h_0
$
- $
h_k = h_0 + \frac{\omega^2 d^2}{2g}
$
- $
h_m = h_0 + \frac{\omega^2 (2d)^2}{2g} = h_0 + \frac{4 \omega^2 d^2}{2g} = h_0 + 2 \frac{\omega^2 d^2}{g}
$
4. Yüksekliklerin Karşılaştırılması
- h_l = h_0 en küçük yükseklik.
- h_k = h_0 + \frac{\omega^2 d^2}{2g} aradaki yükseklik.
- h_m = h_0 + 2 \frac{\omega^2 d^2}{g} en büyük yükseklik.
Dolayısıyla:
h_l < h_k < h_m
5. Seçeneklerle Karşılaştırma
| Seçenek | İfade | Doğruluk Durumu |
|---|---|---|
| A | h_k = h_l = h_m | Yanlış, fark var |
| B | h_l > h_k = h_m | Yanlış, h_l en küçük |
| C | h_k > h_m > h_l | Yanlış, h_m en büyük |
| D | h_l > h_m > h_k | Yanlış, yükseklik artıyor |
| E | h_l < h_k < h_m | Doğru |
6. Özet Tablosu
| Nokta | Dönme Ekseni Uzaklığı r | Yükseklik Formülü | Yükseklik Sırası |
|---|---|---|---|
| L | 0 | h_l = h_0 | En küçük |
| K | d | h_k = h_0 + \frac{\omega^2 d^2}{2g} | Orta |
| M | 2d | h_m = h_0 + 2 \frac{\omega^2 d^2}{g} | En büyük |
Sonuç
Dönme etkisi ile suyun serbest yüzeyi parabolik bir şekilde yükselir. Eksene en yakın noktada (L) yükseklik en düşük, uzakta (M) en yüksek olur. Buna göre;
h_l < h_k < h_m ilişkisi doğrudur ve bu sorunun doğru cevabı E) şıkkıdır.
Eğer istersen formüller ve sıvı dönmesi hakkında daha fazla detaylı açıklama sağlayabilirim ya da farklı örneklerle konuyu pekiştirebilirim. Başarılar! @Mr.Yasin
Formül nereden geliyor?
Question: Formül nereden geliyor? (Where does the formula come from?)
Detailed Explanation of the Formula for Fluid Surface Height in a Rotating Container
When a liquid rotates steadily around a vertical axis, the free surface of the liquid is not flat but takes the shape of a paraboloid. This shape arises due to the balance of forces acting on the fluid elements inside the rotating container.
1. Physical Principles and Forces Involved
- Gravity (g): Pulls the fluid downward.
- Centrifugal Force: Acts outward radially on the fluid elements due to rotation with angular velocity \omega.
- Pressure Gradient: Develops inside the fluid to balance these forces, leading to the curved free surface.
2. Setting up the Problem
Consider a fluid rotating around the vertical z-axis with constant angular velocity \omega.
- r : Radial distance from the axis of rotation.
- z(r) : Height of the fluid surface at radius r.
- Fluid is in equilibrium — no net flow in steady rotation.
3. Derivation of the Shape of the Free Surface
Step 1: Write the acceleration components felt by a fluid particle rotating with angular velocity \omega:
- Radially outward centrifugal acceleration:a_r = \omega^2 r
- Vertically downward gravitational acceleration:a_z = g
Step 2: Apply fluid equilibrium condition (Euler equation) in cylindrical coordinates for a rotating fluid at rest (steady rotation):
The pressure gradient balances the body forces acting on fluid elements:
\frac{1}{\rho} \frac{\partial P}{\partial r} = \omega^2 r \quad ; \quad \frac{1}{\rho} \frac{\partial P}{\partial z} = -g
where
- P : pressure,
- \rho : density of the fluid.
Step 3: Integrate to find pressure inside the fluid:
- Integrating vertically:
P(r, z) = P_0(r) - \rho g z
where P_0(r) is pressure at reference plane z=0.
- Integrating radially:
\frac{\partial P}{\partial r} = \rho \omega^2 r \implies P(r,z) = P_{\text{ref}} - \rho g z + \frac{1}{2} \rho \omega^2 r^2
Step 4: Determine the free surface shape
At the free surface of the fluid, the pressure equals atmospheric pressure P_{\text{atm}} (constant):
P(r, z) = P_{\text{atm}}
Rearranged for the height z(r) of the free surface:
P_{\text{atm}} = P_{\text{ref}} - \rho g z(r) + \frac{1}{2} \rho \omega^2 r^2
Solve for z(r):
z(r) = \frac{P_{\text{ref}} - P_{\text{atm}}}{\rho g} + \frac{\omega^2 r^2}{2g}
Set the constant term as z_0:
\boxed{
z(r) = z_0 + \frac{\omega^2 r^2}{2g}
}
4. Interpretation
- The free surface height z(r) increases quadratically with the radius r, forming a paraboloid.
- z_0 is the height of the fluid surface at the axis (r=0).
- This formula explains why the fluid rises at the outer edges when rotated.
5. Summary Table
| Quantity | Meaning | Formula / Explanation |
|---|---|---|
| \omega | Angular velocity (rad/s) | Given by rotation speed |
| r | Radial distance from rotation axis | r=0 at axis, increases outward |
| g | Gravitational acceleration (9.8 m/s²) | Constant |
| z_0 | Reference height at axis (r=0) | Constant determined by fluid volume |
| z(r) | Height of fluid surface at radius r | z_0 + \frac{\omega^2 r^2}{2 g} |
| Pressure at free surface P | Atmospheric pressure | Constant |
6. Visual Explanation
- Imagine placing water in a bucket and spinning it.
- The surface curves upward at edges, dips near the center.
- This curvature is exactly described by the parabolic equation above.
7. Importance in the Original Problem
Using this formula, you can calculate how the heights of water columns at different radii (r values) relate, as we did before:
- h(r) = h_0 + \frac{\omega^2 r^2}{2g}
- Larger radius → higher water level.
This solves why h_m > h_k > h_l.
Summary
The formula
z(r) = z_0 + \frac{\omega^2 r^2}{2g}
comes from applying the equilibrium equations for fluid pressure under gravity and centrifugal force in a rotating frame, balancing forces to determine the shape of the free surface — a paraboloid.
If you want, I can help you visualize this with a diagram or give example calculations! @Mr.Yasin