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VinnyCee
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Kinetics in Nromal and Tangential coordinates - SPINNING ROD W/ SPOOL
The 2-kg spool fits loosely on the inclined rod for which the coefficient of static friction is [itex]\mu_s\,=\,0.2[/itex]. If the spool is located 0.25 m from A, determine the maximum speed the spool can have so that it does not slip up the rod.
http://img207.imageshack.us/img207/6962/problem13770pa.jpg
Here is what I have for this problem, but it is wrong according to the answer in the text:
[tex]sin\,\theta\,=\,\frac{3}{5}[/tex]
[tex]\theta\,=\,36.9[/tex]
[tex]\sum\,F_x\,=\,N\,cos\,53.1\,-\,f_s\,cos\,36.9\,=\,m\,a_n[/tex]
[tex]\sum\,F_y\,=\,N\,sin\,53.1\,+\,f_s\,sin\,36.9\,-\,m\,g\,=\,m\,a_y[/tex]
[tex]N\,=\,\frac{m\,g}{sin\,53.1\,+\,\mu_s\,sin\,36.9}[/tex]
[tex]N\,=\,21.31\,N[/tex]
Then I use another version of the [itex]F_x[/itex] equation to solve for [itex]v_{max}[/itex]:
[tex]N\,cos53.1\,-\,N\,\mu_s\,cos\,36.9\,=\,m\,\frac{v_{max}^2}{\rho}[/tex]
[tex]v_{max}^2\,=\,\frac{9.39}{10}[/tex]
[tex]v_{max}\,=\,0.969\,\frac{m}{s}[/tex]
The answer is 1.48 in the text though! Any suggestions?
The 2-kg spool fits loosely on the inclined rod for which the coefficient of static friction is [itex]\mu_s\,=\,0.2[/itex]. If the spool is located 0.25 m from A, determine the maximum speed the spool can have so that it does not slip up the rod.
http://img207.imageshack.us/img207/6962/problem13770pa.jpg
Here is what I have for this problem, but it is wrong according to the answer in the text:
[tex]sin\,\theta\,=\,\frac{3}{5}[/tex]
[tex]\theta\,=\,36.9[/tex]
[tex]\sum\,F_x\,=\,N\,cos\,53.1\,-\,f_s\,cos\,36.9\,=\,m\,a_n[/tex]
[tex]\sum\,F_y\,=\,N\,sin\,53.1\,+\,f_s\,sin\,36.9\,-\,m\,g\,=\,m\,a_y[/tex]
[tex]N\,=\,\frac{m\,g}{sin\,53.1\,+\,\mu_s\,sin\,36.9}[/tex]
[tex]N\,=\,21.31\,N[/tex]
Then I use another version of the [itex]F_x[/itex] equation to solve for [itex]v_{max}[/itex]:
[tex]N\,cos53.1\,-\,N\,\mu_s\,cos\,36.9\,=\,m\,\frac{v_{max}^2}{\rho}[/tex]
[tex]v_{max}^2\,=\,\frac{9.39}{10}[/tex]
[tex]v_{max}\,=\,0.969\,\frac{m}{s}[/tex]
The answer is 1.48 in the text though! Any suggestions?
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