एकसमान निघर्षण को मानते हुए चपटे कॉलर बेयरिंग में संचारित घर्षण बल आघूर्ण निम्न द्वारा दिया जाता है:

(जहां, μ = घर्षण गुणांक, P = बेयरिंग पर कुल प्रणोद, और r1 और r2 कॉलर की आंतरिक और बाहरी त्रिज्या हैं)

This question was previously asked in
MPPGCL JE ME 19 March 2019 Official Paper
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  1. \(T=\frac{1}{2} μ P\left(r_1+r_2\right)\)
  2. \(T=\frac{1}{2} μ P\left(r_1-r_2\right)\)
  3. \(T=\frac{2}{3} \mu P\left(r_1+r_2\right)\)
  4. \(T=\frac{2}{3} \mu P\left(r_1-r_2\right)\)

Answer (Detailed Solution Below)

Option 1 : \(T=\frac{1}{2} μ P\left(r_1+r_2\right)\)
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Detailed Solution

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व्याख्या:

विशेष

घर्षण बल आघूर्ण 

 

एक समान दाब

एक समान निघर्षण

(1) चपटा कीलक बेयरिंग

\(\frac{2}{3}\mu WR\)

\(\frac{1}{2}\mu WR\)

(2) शांकव कीलक बेयरिंग

\(\frac{2}{3}\mu WR\;cosec\;\alpha\)

\(\frac{1}{2}\mu WR\;cosec\;\alpha \)

(3) समलंबाकार शांकव कीलक बेयरिंग

\(\frac{2}{3}\mu W\;cosec\;\alpha \times \left[ {\frac{{r_1^3 - r_2^3}}{{r_1^2 - r_2^2}}} \right]\)

\(\frac{1}{2}\mu W\;cosec\;\alpha \left( {{r_1} + {r_2}} \right)\)

(4) चपटा कॉलर बेयरिंग

\(\frac{2}{3}\mu W\left[ {\frac{{r_1^3 - r_2^3}}{{r_1^2 - r_2^2}}} \right]\)

\(\frac{1}{2}\mu W\left( {{r_1} + {r_2}} \right)\)

(5) एकल डिस्क या पट्टिका क्लच

\(\frac{2}{3}\mu \left[ {\frac{{r_1^3 - r_2^3}}{{r_1^2 - r_2^2}}} \right]\)

 

(6) अस्थायी शंकु

\(\frac{2}{3}\mu W\;cosec\;\alpha \times \left[ {\frac{{r_1^3 - r_2^3}}{{r_1^2 - r_2^2}}} \right]\)

\(\frac{1}{2}\mu W\;cosec\;\alpha \left[ {{r_1} + {r_2}} \right]\)
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