سؤال هام عن السرعة الزاوية

[Dr. Sami]

الأخوة الكرام أعضاء المنتدى
ورد في كتاب ميكانيكا الموائع (تأليف / محمد هشام صديق) طريقة حساب عزم القوى المؤثرة على الجملة التالية:
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M=F1 L - F2 L =(F1 - F2) L
M=1/2 P A V2 (CD1 –CD2) L
M=1/2 π R2 V3 (CD1 - CD2) L
حيث: M - محصلة العزوم المطبقة على الجملة
F1 , F2 - القوى المؤثرة على أنصاف الكرات في الجملة.
L - طول الذراع.
P - ضغط الهواء المطبق على المساحةA المكافئة لنصف الكرة.
V - سرعة الريح.
CD1 , CD2 - معامل الإعاقة لأنصاف الكرات:
CD1=1.42 - التيار يواجه الجانب المقعر.
CD2=0.38 - التيار يواجه الجانب المحدب.
وبالرغم من البحث لم أجد أية علاقة لحساب السرعة الزاوية لهذه الجملة بفرض أن الكتل معروفة.
يرجى من أعضاء المنتدى الكرام المساعدة.

[Dr. Sami]

لم تظهر الصورة ولم أستطع إرفاقها بالمشاركة (أعطى رسالة فشل رفع الملف)
الصورة عبارة نصفي كرة قطر كل منهما R متصلتين بحامل بشكل متعاكس (السطح المقعر لأحدهما مواجه للريح والأخرى سطحها المحدب مواجه للريح.
تدور المجموعة حول محور الحامل.
L - هي طول الذراع (من مركز نصف الكرة إلى محور الدوران)
الجملة تدور نتيجة فرق مقاومة السطحين المقعر والمحدب للهواء.

[Dr. Sami]

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[Dr. Sami]

آسف على التأخير
قمت بتحميل الصورة على الرابط التالي
http://www.m5zn.com/uploads2/2012/1/3/photo/010312150118fpqrpvf2vz75iz.jpg
يرجى ممن لديه معادلة لحساب السرعة الزاوية لهذه الجملة تزويدي بها ولكم مني جزيل الشكر

[azoz20066]

السلام عليكم

اولا باستخدام هذا القانون
http://upload.wikimedia.org/wikipedia/en/math/9/b/e/9be08b9254aaacbc0386b26bf137f2ae.png

حيث أن T هي محصلة العزم الكلي

I هي عزم القصور الذاتي حول محور الدوران

α تسارع الزاوي

فمن خلال تسارع الزاويه تجد سرعة الزاويه

ملاحظه: لست متأكد من الاجابه ولكن حبيت اشاركك بما أملك لعلك تجد فيما كتبت شيئا يفيدك

[ABADY100]

طيب هل تمتلك كتلة الكرتين؟

[Dr. Sami]

الأخوة ABADY100 و azoz20066 أشكركم على مشاركتكم
الأخ ABADY100
يمكنك استخدام الرموز في القوانين اللازمة للحل لأني أرغب في دراسة هذه الجملة بشكلها العام والابتعاد عن الحالات الخاصة وبالتالي يمكنك اعتبار كتلة نصف الكرة المجوفة هي m1 وكتلة الحامل هي m2.
الأخ azoz20066
إن القانون الذي نوهت عنه (لم يظهر في المشاركة وهو أن محصلة العزوم المطبقة تساوي جداء عزم العطالة بالتسارع الزاوي) هو قانون مهم جداً في هذه الحالة. ولكن لم يساعدني في تحديد السرعة الزاوية.
إن هذه الحالة معقدة جداً حسب اعتقادي لذلك شغلت تفكيري جداً.
إن القوانين التي ذكرتها في بداية طرح الموضوع (مع تصحيح بعض الأخطاء التي أعتذر بشدة عنها):
M=F1 L – F2 L =(F1 – F2) L
M=P A (CD1 – CD2) L
M=1/2 ρ A V^2 (CD1 – CD2) L
M=1/2 π ρ R^2 V^2 (CD1 – CD2) L
حيث: ρ - كثافة الهواء
الإشارة ^: تدل على الرفع إلى قوة (أس) مثلاً V^2 هي مربع السرعة.
يمكن تطبيق هذه القوانين فقط عند إقلاع هذه الجملة, لأنه ومع ازدياد السرعة الزاوية وبالتالي ازدياد السرعة الخطية لأنصاف الكرات v سيقل تأثير الهواء على الجهة اليمنى (المقعرة) لأن السرعة النسبية للهواء ستصبح V – v, وبالمقابل سيزداد تأثير الهواء على الجهة اليسرى (المحدبة) لأن السرعة النسبية للهواء ستصبح V + v, وبالتالي فإن العزم المطبق على الجملة سيتناقص. وسيستمر ازدياد السرعة الزاوية وتناقص العزم حتى يتساوى العزم المطبق مع عزم قوى الاحتكاك للمحور وبالتالي تدور الجملة بسرعة زاوية ثابتة (حركة دائرية منتظمة).
والقانون في حالة حركة المجموعة يصبح:
M=1/2 π ρ R^2 ((V – v)^2 CD1 – (V + v)^2 CD2) L