Question
Download Solution PDFद्रव्यमान m का एक कण x-y तल में गति करता है। किसी भी क्षण कण के निर्देशांक x = a cos ωt, y = b sin ωt द्वारा दिए गए हैं, जहाँ a, b और ω स्थिरांक हैं। निर्देशांक प्रणाली के मूल के सापेक्ष कण का कोणीय संवेग क्या होगा?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFसंप्रत्यय:
मूल के सापेक्ष एक कण का कोणीय संवेग निम्न द्वारा दिया जाता है:
\( \vec{L} = \vec{r} \times m\vec{v} \)
जहाँ:
- \( \vec{r} \) स्थिति सदिश है
- \( \vec{v} \) वेग सदिश है
- \( m \) कण का द्रव्यमान है
दिया गया है:
\( x = a \cos(\omega t), \quad y = b \sin(\omega t) \)
स्थिति सदिश: \( \vec{r} = a \cos(\omega t) \hat{i} + b \sin(\omega t) \hat{j} \)
वेग सदिश: \( \vec{v} = \frac{d\vec{r}}{dt} = -a\omega \sin(\omega t) \hat{i} + b\omega \cos(\omega t) \hat{j} \)
गणना:
\( \vec{L} = m(\vec{r} \times \vec{v}) \)
क्रॉस उत्पाद के लिए सारणिक का उपयोग करते हुए:
\( \vec{L} = m \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a \cos(\omega t) & b \sin(\omega t) & 0 \\ -a\omega \sin(\omega t) & b\omega \cos(\omega t) & 0 \end{vmatrix} \)
\( \vec{L} = m[(a \cos(\omega t) \cdot b\omega \cos(\omega t)) - (b \sin(\omega t) \cdot -a\omega \sin(\omega t))] \hat{k} \)
\( \vec{L} = m \cdot ab\omega [\cos^2(\omega t) + \sin^2(\omega t)] \hat{k} \)
\( \vec{L} = ab\omega m \hat{k} \)
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