# Justification for Loop de Loop minimum speed

I was trying to figure out the minimum speed an object would have to travel on a loop not to lose contact with the loop. Setting the centripetal force equal to gravity $$m\frac{v^2}{r} = mg$$ gives $$v = \sqrt{gr}$$ that explanation is valid and makes sense to me but I was wondering why a conservation of energy approach wasn't. Entering the loop with speed $$v$$ and setting Kinetic energy equal to gravitational potential $$0.5mv^2 = mgR$$ gives $$v = \sqrt{2gr}$$ which obviously is not the same. Why is this explanation not correct?

• If you use the search box to search for “loop de loop” you’ll find other questions that may address this. Dec 19, 2019 at 5:17
• @SKDash What is my assumption? Dec 19, 2019 at 5:38
• @JoshZwiebel there are two $v$s--one at the bottom, one at the top..think about it Dec 19, 2019 at 6:34

## 4 Answers

First case :- In the first case(where you used centripetal force), the velocity $$v=\sqrt{gr}$$ is, in fact, the velocity of the object at the top of the loop. To find the velocity of the object at the bottom of the loop, you will need to use energy conservation.

$$\frac{1}{2}m((\sqrt{gr})^2-v_{bottom}^2)=-2mgr \Rightarrow v_{bottom}=\sqrt{5gr}$$

Second case :- In the second case, you assumed the velocity at the topmost point of the loop to be $$0$$. So, by energy conservation,

$$\frac{1}{2}m(0-v_{bottom}^2)=-2mgr \Rightarrow v_{bottom}=\sqrt{4gr}$$

Well, you can see that in the second case, the bottom velocity comes out to be lesser than that in the first case. It is because, the assumption that the velocity at the topmost point being zero is wrong. Imagine it this way: If, somehow, the velocity at the topmost point becomes zero(well, that would never be the case as the object, when thrown with bottom velocity $$=\sqrt{4gr}$$, will surely leave the contact from the loop before reaching the topmost point), then how do you expect the object to complete the loop. The object would stop and gravity would take over and pull the object back to the ground without letting it complete the loop. So you need to throw the object a little faster($$v=\sqrt{5gr}$$) such that it always stays with the loop and does not fall down midway between the loop.

What you have done is, you have taken the initial kinetic energy as $$\frac{1}{2}mv^2$$, then you have taken the change in potential energy to be $$2mgR$$ at the topmost point not$$mgR$$, that was your first mistake. Even then you are having zero velocity at the topmost point, which means the body has no velocity, and hence would just fall down instead of completing the loop.

$$\sqrt{gr}$$ is the minimum speed needed at the top of the loop to maintain circular motion

$$\sqrt{2gh}$$ is the minimum speed needed at the bottom to reach a height of $$h$$. To reach the top of the loop $$h=2r$$, the diameter of the circle.

Conservation of energy can be applied to find the minimum speed needed at the bottom to maintain circular motion. If $$u$$ is the speed at the bottom and $$v$$ is that at the top, then

$$\frac{1}{2}mu^2=\frac{1}{2}mv^2 + 2mgr$$ $$\frac{mv^2}{r}=mg$$

This gives

$$\frac{1}{2}mu^2=\frac{1}{2}mgr+2mgr$$

which leads to

$$u=\sqrt{5gr}$$. This is the minimum speed needed at the bottom for circular motion to occur and should not be confused with other minumum speeds.

To work out the minimum speed, the speed at the top must be 0. The speed at the bottom is the minimum speed that you are looking for; it can be calculated by equating the PE at the top to the KE at the bottom because the energy has been converted from PE to KE when an object travels down a loop.

$$0.5mv^2$$ = $$mgh$$

$$h$$ = $$2r$$

Here you can work out the speed by simple algebraic manipulations.