Answer:
a)[tex]|N|=9.83\: N[/tex] at the top
[tex]|N|=33.37\: N[/tex] at the botton
b) The minimum velocity will be [tex]v=4.43\: m/s[/tex].
Step-by-step explanation:
a) Using the second Newton's law, at the top of the circle we have.
[tex]\Sigma F=ma_{c}[/tex]
The forces at the top are the weight and the normal force.
[tex]W-N=m\frac{v^{2}}{R}[/tex]
[tex]mg-N=m\frac{v^{2}}{R}[/tex]
[tex]N=mg-(m\frac{v^{2}}{R})[/tex]
[tex]N=1.2*9.81-(1.2\frac{6^{2}}{2})[/tex]
[tex]N=-9.83\: N[/tex]
[tex]|N|=9.83\: N[/tex]
At the botton of the circle we have:
[tex]N-W=m\frac{v^{2}}{R}[/tex]
[tex]N-mg=m\frac{v^{2}}{R}[/tex]
[tex]N=1.2*9.81+(1.2\frac{6^{2}}{2})[/tex]
[tex]|N|=33.37\: N[/tex]
b) If we do the normal force equal to zero we can find the minimum velocity, which means:
[tex]W-0=m\frac{v^{2}}{R}[/tex]
[tex]mg=m\frac{v^{2}}{R}[/tex]
[tex]v=\sqrt{Rg}[/tex]
[tex]v=\sqrt{2*9.81}[/tex]
Therefore, the minimum velocity will be [tex]v=4.43\: m/s[/tex].
I hope it helps you!
