Question

Please help me go over the practice assignment that's attached.

Answer 1

To solve the problem first, we need the acceleration, in this case, the centripetal acceleration

`a=(v^2)/(R)`

1. v=3m/s , R=0.7m

`a_c=(3^2)/(0.7)=(12.86m)/(s^2)`

now the sum of forces, remember the centripetal force always goes to the center. The forces that interact in this point are weight (goes down), centripetal (goes down), and tension which has no defined direction but I'm going to assume it goes up.

`\begin{gathered} \sum ^(\infty)_(n\mathop=0)Fy=0 \\ 0=T-mg-ma_c \\ T=2\cdot9.81+2\cdot12.86 \\ T=45.34N\text{ going up} \end{gathered}`

2. v=5m/s , R=0.7m

`a_c=(5^2)/(.7)=(35.71m)/(s^2)`

now the sum of forces, remember the centripetal force always goes to the center. The forces that interact in this point are weight (goes down), centripetal (goes up), and tension which has no defined direction but I'm going to assume it goes down.

`\begin{gathered} \sum ^(\infty)_(n\mathop=0)Fy=0 \\ 0=ma_c-mg-T \\ T=2\cdot35.71-2\cdot9.81 \\ T=51.8N\text{ going down} \end{gathered}`

This is for 1.

For 2.