Zhadnyy, boa tarde !
- Screenshot_20200703-124519~2.png (49.31 KiB) Exibido 714 vezes
[tex3]I.[/tex3]
Velocidade do corpo quando perde contato com o trilho:
[tex3]F_{cp}=\frac{mv^2}R[/tex3]
[tex3]mg\sen\alpha=\frac{mv^2}R[/tex3]
[tex3]v=\sqrt{gR\sen\alpha} \ \rightarrow \ \begin{cases} v_x=v\sen\alpha \ \implies \ v_x =\sen\alpha\sqrt{gR\sen\alpha} \\ v_y=v\cos\alpha \ \implies \ v_y=\cos\alpha\sqrt{gR\sen\alpha}\end{cases}[/tex3]
[tex3]II.[/tex3]
Movimento no eixo [tex3]x[/tex3]
:
[tex3]S=S_0+v_x\cdot t[/tex3]
[tex3]R\cos\alpha=\sen\alpha\sqrt{gR\sen\alpha}\cdot t[/tex3]
[tex3]\boxed{t=\frac{R\cos\alpha}{\sen\alpha\sqrt{gR\sen\alpha}}}[/tex3]
[tex3]III.[/tex3]
Movimento no eixo [tex3]y[/tex3]
:
[tex3]S=S_0+v_y\cdot t-\frac{gt^2}2[/tex3]
[tex3]0=R\sen\alpha+\cos\alpha\sqrt{gR\sen\alpha}\cdot \frac{R\cos\alpha}{\sen\alpha\sqrt{gR\sen\alpha}}-\frac{gR^2\cos^2\alpha}{2gR\sen^3\alpha}[/tex3]
[tex3]0=\sen\alpha+\frac{\cos^2\alpha}{\sen\alpha}-\frac{\cos^2\alpha}{2\sen^3\alpha}[/tex3]
[tex3]0=2\sen^4\alpha+2\sen^2\alpha\cos^2\alpha-\cos^2\alpha[/tex3]
[tex3]2\sen^4\alpha+2\sen^2\alpha\(1-\sen^2\alpha\)-\(1-\sen^2\alpha\)=0[/tex3]
[tex3]2\sen^4\alpha-2\sen^4\alpha+2\sen^2\alpha-1+\sen^2\alpha=0[/tex3]
[tex3]3\sen^2\alpha=1 \ \implies \ \boxed{\sen\alpha=\frac{\sqrt 3}3}[/tex3]
[tex3]IV.[/tex3]
Conservação de energia:
[tex3]E_{M_0}=E_{M_f}[/tex3]
[tex3]mgH=\frac{mv^2}2+mg\(R+R\sen\alpha\)[/tex3]
[tex3]gH=\frac{gR\sen\alpha}2+gR\(1+\sen\alpha\)[/tex3]
[tex3]H=R\(1+\frac{3\sen\alpha}2\)[/tex3]
[tex3]H=\(1+\frac 32\cdot \frac{\sqrt 3}3\)R[/tex3]
[tex3]\boxed{\boxed{H=\(1+\frac{\sqrt3}2\)R}}[/tex3]