In vertical circular motion while finding the minimum velocity at the bottom-most point for looping the loop, why do we take the tension at the topmost point to be zero?
Vertical Circular Motion – Calculating Minimum Velocity Required
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Related Solutions
What is the configuration needed for a circular-motion?
The answer is, there must be an inward(towards the center) force perpendicular to the instantaneous velocity. The minimum velocity required for the bob initially to loop the whole loop is $\sqrt{5gR}$. In this situation, the tension at the topmost point becomes zero, but since there is velocity, which is $\sqrt{Rg}$ & there is inward force, which is the weight $mg$; this matches up the configuration of circular-motion & hence continues to complete the circle.
Now, if the initial velocity is less than $\sqrt{5gR}$, then the bob will have attained zero velocity before reaching the top. Now, using this velocity, we can find $h_1 \; , \; h_2$, the heights where the tension becomes zero & the height where the tangential instantaneous velocity of the bob becomes zero respectively. Using kinematics equation, we find $$h_1 = \frac{v_\text{initial}^2 + Rg}{3g} \\ h_2 = \frac{v_\text{initial}^2}{2g}$$.
Now, the case you mentioned can be described as $h_1 \lt h_2$ i.e. the tension becomes zero before velocity becomes & this implies certain velocity does exist when tension becomes zero.
The gravitational force never does point inwards i.e. towards the center of the circular trajectory. The component $mg\sin \theta$ opposes the tangential velocity & remains collinear with the instantaneous velocity vector but in the opposite direction; $mg\cos \theta$, on the other hand, points outward i.e. away from the center just opposite to the centripetal force i.e. tension. You can see in the pic$^1$ how the components of gravitational force are specified; no one is acting inwards i.e. towards the center of the circle.
When tension becomes zero, velocity still exists & $mg\cos\theta$ remains perpendicular to it & is pointing inward to some point; hence it matches the configuration of circular-motion. The bob thus breaks its old circular trajectory & undergoes circular motion around a new loop where the centripetal force is $mg\cos\theta$. But it can't undergo a pure circular motion as the bob is retarded by the opposing force $mg\sin\theta$ & thus it follows a parabolic trajectory.
So, in a nut-shell, it is gravity which lets the bob to move away from the trajectory when tension becomes zero.
$^1$ Courtsey: Understanding Physics vol.I by D.C.Pandey.
If the tension is equal zero at some point below the uppermost level, the tension at higher points would need to be negative for the circular motion to continue. Of course, the tension cannot be negative, instead, the rope would become loose and the object would fall.
Best Answer
To find the minimum velocity at the bottom-most point, we find the minimum velocity at the uppermost point. This minimum velocity is the one such that the centripetal force is equal to the force of gravity. Any lower and gravity will pull the object down and out of the loop, any higher and a faster velocity would be required to generate it (thus, not a minimum). Because this is the case, we set the tension to be zero because gravity can account for the entire centripetal force. Using this minimum velocity at the topmost point, we can use conservation of energy to find the minimum velocity at the bottom-most point.