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For part (a) which rod should I take the Aluminum or the brass and why ? 

 

 

physics
 Nov 1, 2015

Best Answer 

 #1
avatar+26226 
+45

 

For part (a) which rod should I take the Aluminum or the brass and why ? 

 

Shaft\(_{ \mathbf{\text{aluminum} }}\)   CB:

 

\(\begin{array}{lcl} c=\frac12 d = 0.006\ \text{m} \qquad L=0.300\ \text{m} \qquad G= 26\cdot 10^9\ \text{Pa} \qquad T=100\ \text{N}\cdot \text{m}\\\\ J=\frac{\pi}{2}c^4 = \frac{\pi}{2}\cdot (0.006^4)\ \text{m}^4 = 2.0357520\cdot 10^{-9}\ \text{m}^4 \\\\ \varphi_{_{\text{CB}}} = \frac{T\cdot L}{G\cdot J} = \frac{100\cdot 0.300\ \text{N}\cdot \text{m}^2} {26\cdot 10^9\cdot 2.0357520\cdot 10^{-9}\ \text{Pa}\cdot\text{m}^4} = \frac{30}{52.929553}\ \text{rad} =0.5667911079\ \text{rad} \end{array}\)

 

 

Shaft\(_{ \mathbf{ \text{brass}}}\)  BA:

 

\(\begin{array}{lcl} c=\frac12 d = 0.006\ \text{m} \qquad L=0.200\ \text{m} \qquad G= 39\cdot 10^9\ \text{Pa} \qquad T=100\ \text{N}\cdot \text{m}\\\\ J=\frac{\pi}{2}c^4 = \frac{\pi}{2}\cdot (0.006^4)\ \text{m}^4 = 2.0357520\cdot 10^{-9}\ \text{m}^4 \\\\ \varphi_{_{\text{BA}}} = \frac{T\cdot L}{G\cdot J} = \frac{100\cdot 0.200\ \text{N}\cdot \text{m}^2} {39\cdot 10^9\cdot 2.0357520\cdot 10^{-9}\ \text{Pa}\cdot\text{m}^4} = \frac{20}{79.3943295415}\ \text{rad} =0.2519071591\ \text{rad} \end{array}\)

 

\(\begin{array}{lrcl} \text{answers:}\\ (a) & \varphi_{_{\text{B}}}&=& \varphi_{_{\text{BA}}}\\ &&=& 0.2519071591\ \text{rad} \\ &&=& 14.4332170431^{\circ}\\\\ \hline (b)& \varphi_{_{\text{C}}} &=& \varphi_{_{\text{BA}}} + \varphi_{_{\text{CB}}}\\ &&=& 0.2519071591\ \text{rad}+0.5667911079\ \text{rad}\\ &&=&0.8186982670\ \text{rad}\\ &&=& 46.9079553913^{\circ} \end{array} \)

 

For part (a) which rod should I take the Aluminum or the brass and why ?

\( \varphi_{_{ \text{B}_{\text{brass}} }} < \varphi_{_{\text{B}_{\text{aluminum}} }} \)

because

\(G_{\text{brass}} > G_{\text{aluminum}} \)

laugh

 Nov 2, 2015
 #1
avatar+26226 
+45
Best Answer

 

For part (a) which rod should I take the Aluminum or the brass and why ? 

 

Shaft\(_{ \mathbf{\text{aluminum} }}\)   CB:

 

\(\begin{array}{lcl} c=\frac12 d = 0.006\ \text{m} \qquad L=0.300\ \text{m} \qquad G= 26\cdot 10^9\ \text{Pa} \qquad T=100\ \text{N}\cdot \text{m}\\\\ J=\frac{\pi}{2}c^4 = \frac{\pi}{2}\cdot (0.006^4)\ \text{m}^4 = 2.0357520\cdot 10^{-9}\ \text{m}^4 \\\\ \varphi_{_{\text{CB}}} = \frac{T\cdot L}{G\cdot J} = \frac{100\cdot 0.300\ \text{N}\cdot \text{m}^2} {26\cdot 10^9\cdot 2.0357520\cdot 10^{-9}\ \text{Pa}\cdot\text{m}^4} = \frac{30}{52.929553}\ \text{rad} =0.5667911079\ \text{rad} \end{array}\)

 

 

Shaft\(_{ \mathbf{ \text{brass}}}\)  BA:

 

\(\begin{array}{lcl} c=\frac12 d = 0.006\ \text{m} \qquad L=0.200\ \text{m} \qquad G= 39\cdot 10^9\ \text{Pa} \qquad T=100\ \text{N}\cdot \text{m}\\\\ J=\frac{\pi}{2}c^4 = \frac{\pi}{2}\cdot (0.006^4)\ \text{m}^4 = 2.0357520\cdot 10^{-9}\ \text{m}^4 \\\\ \varphi_{_{\text{BA}}} = \frac{T\cdot L}{G\cdot J} = \frac{100\cdot 0.200\ \text{N}\cdot \text{m}^2} {39\cdot 10^9\cdot 2.0357520\cdot 10^{-9}\ \text{Pa}\cdot\text{m}^4} = \frac{20}{79.3943295415}\ \text{rad} =0.2519071591\ \text{rad} \end{array}\)

 

\(\begin{array}{lrcl} \text{answers:}\\ (a) & \varphi_{_{\text{B}}}&=& \varphi_{_{\text{BA}}}\\ &&=& 0.2519071591\ \text{rad} \\ &&=& 14.4332170431^{\circ}\\\\ \hline (b)& \varphi_{_{\text{C}}} &=& \varphi_{_{\text{BA}}} + \varphi_{_{\text{CB}}}\\ &&=& 0.2519071591\ \text{rad}+0.5667911079\ \text{rad}\\ &&=&0.8186982670\ \text{rad}\\ &&=& 46.9079553913^{\circ} \end{array} \)

 

For part (a) which rod should I take the Aluminum or the brass and why ?

\( \varphi_{_{ \text{B}_{\text{brass}} }} < \varphi_{_{\text{B}_{\text{aluminum}} }} \)

because

\(G_{\text{brass}} > G_{\text{aluminum}} \)

laugh

heureka Nov 2, 2015
 #2
avatar+1832 
+5

Thank You

 Nov 5, 2015
 #3
avatar+115747 
+5

WOW, that looks impressive Heureka    laughlaughlaugh

 Nov 5, 2015

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