Explanation
Step 1
Draw
so
a)centripetal force:
the centripetal force is given by.
[tex]\begin{gathered} F=ma \\ F=m\frac{v^2}{r} \\ \text{where } \\ F_{C\text{ }}\text{ is the centripetal force} \\ m\text{ is the mass } \\ v\text{ is the velocty } \\ r\text{ is the radius} \end{gathered}[/tex]now, replace
[tex]\begin{gathered} F=m\frac{v^2}{r} \\ F=1.86\text{ kg }\frac{(\text{ 5.27 }\frac{m}{s})^2}{1.29\text{ m}} \\ F=40.044\text{ Newtons} \end{gathered}[/tex]so, the centripetal force is 40.0446 Newtons
b) What is the tension in the cord when the ball is at the bottom of its path?
to find the tension in bottom, we need to add the weigth of the ball,so
[tex]\begin{gathered} \text{weigth}=\text{ mass}\cdot accelofgravity \\ w=mg \end{gathered}[/tex]hence, the tension would be
[tex]\begin{gathered} T_{bottom}=m\frac{v^2}{r}+mg \\ \end{gathered}[/tex]replace
[tex]\begin{gathered} T_{bottom}=m\frac{v^2}{r}+mg \\ T_{bottom}=40.044\text{ N+(1.86 kg}\cdot9.81\text{ }\frac{\text{m}}{s^2}) \\ T_{bottom}=40.044\text{ N+18.2466 N} \\ T_{bottom}=58.291\text{ N} \end{gathered}[/tex]c)What is the tension in the cord when the ball is at the top of its path?
to find the tension in the top we need to subtract the weigth, so
[tex]\begin{gathered} T_{\text{top}}=m\frac{v^2}{r}-mg \\ replace \\ T_{\text{top}}=40.044\text{ N-18.2466 N} \\ T_{\text{top}}=21.79\text{ Newtons} \end{gathered}[/tex]I hope this helps you
