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assume that a Cylinder with radius [tex]\[<br /> R<br /> \][/tex] , proper mass [tex]\[<br /> M_0 <br /> \][/tex] and height [tex]\[<br /> h<br /> \][/tex] which is rotating at a constant angular speed [tex]\[<br /> \omega <br /> \][/tex]
In order to calculate the relativistic mass we use the proper mass element to calculate the relativistic mass element , so :
[tex]\[<br /> dM = \frac{{dM_0 }}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }}<br /> \][/tex]
But [tex]\[<br /> dM_0 = \rho _0 dV_0 <br /> \][/tex] where [tex]\[<br /> \rho _0 <br /> \][/tex] is the proper mass density and [tex]\[<br /> dV_0 <br /> \][/tex] is the proper volume element . so :
[tex]\[<br /> \begin{array}{l}<br /> dM = \frac{{\rho _0 dV}}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }} \\ <br /> M = \int\limits_V {\frac{{\rho _0 dV}}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }}} = \int\limits_0^R {\int\limits_0^{2\pi } {\int\limits_0^h {\frac{{\rho _0 rdrd\phi dz}}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }}} } } = 2\pi h\rho _0 \int\limits_0^R {\frac{{rdr}}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }}} \\ <br /> but:v = \omega r \\ <br /> M = 2\pi h\rho _0 \int\limits_0^R {\frac{{rdr}}{{\sqrt {1 - \frac{{\omega ^2 }}{{c^2 }}r^2 } }}} \\ <br /> \end{array}<br /> \][/tex]
Now , make the substitution
[tex]\[<br /> u = 1 - \frac{{\omega ^2 }}{{c^2 }}r^2 \Rightarrow du = - 2\frac{{\omega ^2 }}{{c^2 }}rdr \Rightarrow 2rdr = - \frac{{c^2 }}{{\omega ^2 }}du<br /> \][/tex]
so :
[tex]\[<br /> \begin{array}{l}<br /> M = - \frac{{\pi h\rho _0 c^2 }}{{\omega ^2 }}\int\limits_1^{1 - \left( {\frac{{\omega R}}{c}} \right)^2 } {\frac{{du}}{{\sqrt u }}} = - \frac{{2\pi h\rho _0 c^2 }}{{\omega ^2 }}\left[ {\sqrt u } \right]_1^{1 - \left( {\frac{{\omega R}}{c}} \right)^2 } = \frac{{2\pi h\rho _0 c^2 }}{{\omega ^2 }}\left( {1 - \sqrt {1 - \left( {\frac{{\omega R}}{c}} \right)^2 } } \right) \\ <br /> but:M_0 = \rho _0 V_0 = \pi R^2 h\rho _0 \\ <br /> M = \frac{{2M_0 c^2 }}{{R^2 \omega ^2 }}\left( {1 - \sqrt {1 - \left( {\frac{{\omega R}}{c}} \right)^2 } } \right) \\ <br /> \end{array}<br /> \][/tex]
now , there is something make me confused in this equation . If we put [tex]\[<br /> \omega R = c<br /> \][/tex] we find that the relativistic mass is [tex]\[<br /> M = 2M_0 <br /> \][/tex] . How it can be ?
I know that any thing has a v = c it's mass goes to infinity .
Again , How it can be ?
Thanks
assume that a Cylinder with radius [tex]\[<br /> R<br /> \][/tex] , proper mass [tex]\[<br /> M_0 <br /> \][/tex] and height [tex]\[<br /> h<br /> \][/tex] which is rotating at a constant angular speed [tex]\[<br /> \omega <br /> \][/tex]
In order to calculate the relativistic mass we use the proper mass element to calculate the relativistic mass element , so :
[tex]\[<br /> dM = \frac{{dM_0 }}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }}<br /> \][/tex]
But [tex]\[<br /> dM_0 = \rho _0 dV_0 <br /> \][/tex] where [tex]\[<br /> \rho _0 <br /> \][/tex] is the proper mass density and [tex]\[<br /> dV_0 <br /> \][/tex] is the proper volume element . so :
[tex]\[<br /> \begin{array}{l}<br /> dM = \frac{{\rho _0 dV}}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }} \\ <br /> M = \int\limits_V {\frac{{\rho _0 dV}}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }}} = \int\limits_0^R {\int\limits_0^{2\pi } {\int\limits_0^h {\frac{{\rho _0 rdrd\phi dz}}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }}} } } = 2\pi h\rho _0 \int\limits_0^R {\frac{{rdr}}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }}} \\ <br /> but:v = \omega r \\ <br /> M = 2\pi h\rho _0 \int\limits_0^R {\frac{{rdr}}{{\sqrt {1 - \frac{{\omega ^2 }}{{c^2 }}r^2 } }}} \\ <br /> \end{array}<br /> \][/tex]
Now , make the substitution
[tex]\[<br /> u = 1 - \frac{{\omega ^2 }}{{c^2 }}r^2 \Rightarrow du = - 2\frac{{\omega ^2 }}{{c^2 }}rdr \Rightarrow 2rdr = - \frac{{c^2 }}{{\omega ^2 }}du<br /> \][/tex]
so :
[tex]\[<br /> \begin{array}{l}<br /> M = - \frac{{\pi h\rho _0 c^2 }}{{\omega ^2 }}\int\limits_1^{1 - \left( {\frac{{\omega R}}{c}} \right)^2 } {\frac{{du}}{{\sqrt u }}} = - \frac{{2\pi h\rho _0 c^2 }}{{\omega ^2 }}\left[ {\sqrt u } \right]_1^{1 - \left( {\frac{{\omega R}}{c}} \right)^2 } = \frac{{2\pi h\rho _0 c^2 }}{{\omega ^2 }}\left( {1 - \sqrt {1 - \left( {\frac{{\omega R}}{c}} \right)^2 } } \right) \\ <br /> but:M_0 = \rho _0 V_0 = \pi R^2 h\rho _0 \\ <br /> M = \frac{{2M_0 c^2 }}{{R^2 \omega ^2 }}\left( {1 - \sqrt {1 - \left( {\frac{{\omega R}}{c}} \right)^2 } } \right) \\ <br /> \end{array}<br /> \][/tex]
now , there is something make me confused in this equation . If we put [tex]\[<br /> \omega R = c<br /> \][/tex] we find that the relativistic mass is [tex]\[<br /> M = 2M_0 <br /> \][/tex] . How it can be ?
I know that any thing has a v = c it's mass goes to infinity .
Again , How it can be ?
Thanks