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Eli
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The "centripetal force" on a small mass element is

$$dF=dm\,\omega^2 \,x\\ \text{with}\quad dm=A(x)\,\rho\,dx=A\,\rho\,dx $$$$F+dF-F-dm\,\omega^2\,x=0\quad\Rightarrow\\dF=dm\,\omega^2 \,x\\ \text{with}\quad dm=A(x)\,\rho\,dx=A\,\rho\,dx $$

you obtain

$$F=A\,\rho\,\omega^2\int x\,dx=\frac 12 A\,\rho\,\omega^2\,x^2+c\\ \text{with}\quad F(L/2)=0\quad\Rightarrow c=-\frac 18\,A\rho\,{\omega}^{2}L^2$$

thus $$F(x)=\frac 12\,A\,\rho\,\omega^2\left(x^2-\frac{L^2}{4}\right) \quad, 0\le x\le \frac L2$$

you can obtain the angular velocity $~\omega~$ from the EOM

$$I_{\text{CM}}\,\dot\omega=\tau\quad,I_{\text{CM}}=\frac{m\,L^2}{12}$$

enter image description here

The "centripetal force" on a small mass element is

$$dF=dm\,\omega^2 \,x\\ \text{with}\quad dm=A(x)\,\rho\,dx=A\,\rho\,dx $$

you obtain

$$F=A\,\rho\,\omega^2\int x\,dx=\frac 12 A\,\rho\,\omega^2\,x^2+c\\ \text{with}\quad F(L/2)=0\quad\Rightarrow c=-\frac 18\,A\rho\,{\omega}^{2}L^2$$

thus $$F(x)=\frac 12\,A\,\rho\,\omega^2\left(x^2-\frac{L^2}{4}\right) \quad, 0\le x\le \frac L2$$

you can obtain the angular velocity $~\omega~$ from the EOM

$$I_{\text{CM}}\,\dot\omega=\tau\quad,I_{\text{CM}}=\frac{m\,L^2}{12}$$

enter image description here

The "centripetal force" on a small mass element is

$$F+dF-F-dm\,\omega^2\,x=0\quad\Rightarrow\\dF=dm\,\omega^2 \,x\\ \text{with}\quad dm=A(x)\,\rho\,dx=A\,\rho\,dx $$

you obtain

$$F=A\,\rho\,\omega^2\int x\,dx=\frac 12 A\,\rho\,\omega^2\,x^2+c\\ \text{with}\quad F(L/2)=0\quad\Rightarrow c=-\frac 18\,A\rho\,{\omega}^{2}L^2$$

thus $$F(x)=\frac 12\,A\,\rho\,\omega^2\left(x^2-\frac{L^2}{4}\right) \quad, 0\le x\le \frac L2$$

you can obtain the angular velocity $~\omega~$ from the EOM

$$I_{\text{CM}}\,\dot\omega=\tau\quad,I_{\text{CM}}=\frac{m\,L^2}{12}$$

added 9 characters in body
Source Link
Eli
Eli
  • 12.8k
  • 2
  • 11
  • 31

enter image description here

The "centripetal force" on a small mass element is

$$dF=dm\,\omega^2 \,x\\ \text{with}\quad dm=A(x)\,\rho\,dx=A\,\rho\,dx $$

you obtain

$$F=A\,\rho\,\omega^2\int x\,dx=\frac 12 A\,\rho\,\omega^2\,x^2+c\\ \text{with}\quad F(L/2)=0\quad\Rightarrow c=-\frac 14\,A\rho\,{\omega}^{2}L$$$$F=A\,\rho\,\omega^2\int x\,dx=\frac 12 A\,\rho\,\omega^2\,x^2+c\\ \text{with}\quad F(L/2)=0\quad\Rightarrow c=-\frac 18\,A\rho\,{\omega}^{2}L^2$$

thus $$F(x)=\frac 12\,A\,\rho\,\omega^2\left(x-\frac L2\right) \quad, 0\le x\le \frac L2$$$$F(x)=\frac 12\,A\,\rho\,\omega^2\left(x^2-\frac{L^2}{4}\right) \quad, 0\le x\le \frac L2$$

you can obtain the angular velocity $~\omega~$ from the EOM

$$I_{\text{CM}}\,\dot\omega=\tau\quad,I_{\text{CM}}=\frac{m\,L^2}{12}$$

enter image description here

The "centripetal force" on a small mass element is

$$dF=dm\,\omega^2 \,x\\ \text{with}\quad dm=A(x)\,\rho\,dx=A\,\rho\,dx $$

you obtain

$$F=A\,\rho\,\omega^2\int x\,dx=\frac 12 A\,\rho\,\omega^2\,x^2+c\\ \text{with}\quad F(L/2)=0\quad\Rightarrow c=-\frac 14\,A\rho\,{\omega}^{2}L$$

thus $$F(x)=\frac 12\,A\,\rho\,\omega^2\left(x-\frac L2\right) \quad, 0\le x\le \frac L2$$

you can obtain the angular velocity $~\omega~$ from the EOM

$$I_{\text{CM}}\,\dot\omega=\tau\quad,I_{\text{CM}}=\frac{m\,L^2}{12}$$

enter image description here

The "centripetal force" on a small mass element is

$$dF=dm\,\omega^2 \,x\\ \text{with}\quad dm=A(x)\,\rho\,dx=A\,\rho\,dx $$

you obtain

$$F=A\,\rho\,\omega^2\int x\,dx=\frac 12 A\,\rho\,\omega^2\,x^2+c\\ \text{with}\quad F(L/2)=0\quad\Rightarrow c=-\frac 18\,A\rho\,{\omega}^{2}L^2$$

thus $$F(x)=\frac 12\,A\,\rho\,\omega^2\left(x^2-\frac{L^2}{4}\right) \quad, 0\le x\le \frac L2$$

you can obtain the angular velocity $~\omega~$ from the EOM

$$I_{\text{CM}}\,\dot\omega=\tau\quad,I_{\text{CM}}=\frac{m\,L^2}{12}$$

Post Deleted by Eli
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Eli
Eli
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enter image description hereenter image description here

The "centripetal force" on a small mass element is

$$dF=dm\,\omega^2 \,x\\ \text{with}\quad dm=A(x)\,\rho\,dx=A\,\rho\,dx $$

you obtain

$$F=A\,\rho\,\omega^2\int x\,dx=\frac 12 A\,\rho\,\omega^2\,x^2+c\\ \text{with}\quad F(L/2)=0\quad\Rightarrow c=-\frac 14\,A\rho\,{\omega}^{2}L$$

thus $$F(x)=\frac 12\,A\,\rho\,\omega^2\left(x-\frac L2\right) \quad, 0\le x\le \frac L2$$

you can obtain the angular velocity $~\omega~$ from the EOM

$$I_{\text{CM}}\,\dot\omega=\tau\quad,I_{\text{CM}}=\frac{m\,L^2}{12}$$

enter image description here

The "centripetal force" on a small mass element is

$$dF=dm\,\omega^2 \,x\\ \text{with}\quad dm=A(x)\,\rho\,dx=A\,\rho\,dx $$

you obtain

$$F=A\,\rho\,\omega^2\int x\,dx=\frac 12 A\,\rho\,\omega^2\,x^2+c\\ \text{with}\quad F(L/2)=0\quad\Rightarrow c=-\frac 14\,A\rho\,{\omega}^{2}L$$

thus $$F(x)=\frac 12\,A\,\rho\,\omega^2\left(x-\frac L2\right) \quad, 0\le x\le \frac L2$$

you can obtain the angular velocity $~\omega~$ from the EOM

$$I_{\text{CM}}\,\dot\omega=\tau\quad,I_{\text{CM}}=\frac{m\,L^2}{12}$$

enter image description here

The "centripetal force" on a small mass element is

$$dF=dm\,\omega^2 \,x\\ \text{with}\quad dm=A(x)\,\rho\,dx=A\,\rho\,dx $$

you obtain

$$F=A\,\rho\,\omega^2\int x\,dx=\frac 12 A\,\rho\,\omega^2\,x^2+c\\ \text{with}\quad F(L/2)=0\quad\Rightarrow c=-\frac 14\,A\rho\,{\omega}^{2}L$$

thus $$F(x)=\frac 12\,A\,\rho\,\omega^2\left(x-\frac L2\right) \quad, 0\le x\le \frac L2$$

you can obtain the angular velocity $~\omega~$ from the EOM

$$I_{\text{CM}}\,\dot\omega=\tau\quad,I_{\text{CM}}=\frac{m\,L^2}{12}$$

added 89 characters in body
Source Link
Eli
Eli
  • 12.8k
  • 2
  • 11
  • 31
Source Link
Eli
Eli
  • 12.8k
  • 2
  • 11
  • 31