# Is the force on the bob of a pendulum with respect to ground 0 at the bottom?

Kind of a silly question, but imagine there's a pendulum swinging that has some fixed length l and it's at some angle $$\theta$$, then with respect to us, the forces on the bob are: $$F_{rad} = mg \cos(\theta) - T$$ where $$F_{rad}$$ is radial component of net force and T is tension in the string, and $$F_{Tan} = mg\sin(\theta)$$ where $$F_{Tan}$$ is tangential component of the net force on Bob. When the bob is at its lowest point, $$\theta = 0$$ and so the net force on bob is 0, but isn't there always a net force on bob that causes change in direction of velocity to make sure it doesn't move in a straight line? Or does that force only exist in bob's frame of reference?

• Are you sure that $T=mg$ at the bottom? May 15, 2021 at 10:22
• Thanks for replying. I think that is my central problem, we say that T - mg = F_c where F_c is centripetal force but isn't centripetal force just the tension in the string? May 15, 2021 at 10:35
• I'm thinking Tension in the string is greater than mg, but I can't say why would that be the case? Isn't tension just the reactive force? May 15, 2021 at 10:41
• Replace string with a spring then think about it. I think you should get it. May 15, 2021 at 11:31
• I think you're saying the string gets stretched which would make sense but don't we assume that the length of the string is constant? May 16, 2021 at 5:44

No, tension in the string at lowest point can be found from Newton's Second law: $$\mathbf {F_{net}}=m\mathbf a$$ At lowest point, due to circular motion, we will have radial acceleration $$\frac {v^2}{r}$$. So, we have: $$T-mg=\frac {mv^2}{r}$$ If you know the initial value of $$\theta$$, you can also find out the value of $$T$$ in terms of $$\theta$$ by making use of energy conservation to find $$v$$ at the lowest point.