Vertical Circles and Non-Uniform Circular Motion

[MechaMZ]

Homework Statement

If we wanted to calculate the minimum or critical velocity needed for the block to just be able to pass through the top of the circle without the rope sagging then we would start by letting the tension in the rope approaches zero.

http://img147.imageshack.us/img147/6826/3c6634a694b3434387bd810.gif http://img147.imageshack.us/img147/7402/3c46cefd59484d479f07621.gif http://img147.imageshack.us/img147/4192/5ee60dbd561d4cca96b54d5.gif

why must we letting the tension in the rope approaches zero without the rope sagging? what is the force support the object goes to the top without pulling down by gravitation force? was it because of the Newton first law, so the object wanted to remain straight line but pulling towards to the centre by centripetal force?

http://img225.imageshack.us/img225/4176/p621.gif

At point B, why the normal reaction force is or need to be zero? isn't the mg is acting on the track?
will it due to the inertia of the roller-coaster tends to go straight yet pulling back the centripetal force,mg. hence, the contact force to the track is reduced and almost zero since the roller-coaster is really fast.

Hope somebody could answer these to me, thank you

 

[Andrew Mason]

You can answer your questions by drawing a free body diagram for each of the moving objects (the block / roller coaster rider). The forces acting on the block/rider are gravity and the mechanical force of the rope/car. Since gravity cannot be reduced, the minimum force on the object would occur when the mechanical force is at a minimum.

In the rope case, analyse the forces on the object at the top of the loop (let the positive direction be down):

$$\vec{F} = \vec{T}+ \vec{F_g} = \vec{T} + m\vec{g} = m\vec{a}$$

where a is the net acceleration. What is the net acceleration and net force (hint: it is a function of v, the tangential speed)?Now, what is the value of v if T = 0 and the object is still traveling in a circle?

What happens if v is less than this speed? Does the object still travel in a circle?

AM

 

[MechaMZ]

i know how to do it but i don't understand why to do it in this way. why at the maximum point, the tension is zero? and happens the same while the roller-coaster at the peak point, the normal reaction is zero as well, even the roller-coaster is not upside-down.

are all these due to inertia by Newton second law? object tends to remain their original state, so lesser contact force to the track or rope.

 

[MechaMZ]

is my assumption right or wrong, could somebody please correct me =)

 

[Andrew Mason]

i know how to do it but i don't understand why to do it in this way. why at the maximum point, the tension is zero? and happens the same while the roller-coaster at the peak point, the normal reaction is zero as well, even the roller-coaster is not upside-down.

are all these due to inertia by Newton second law? object tends to remain their original state, so lesser contact force to the track or rope.

Since the object goes in a circle, the Tension + mg will always equal mv^2/R. The tension cannot go below 0 and mg cannot change, so the minimum force is mg. This means that the minimum circular speed is $\sqrt{gR}$. If v is less than this what happens at the top?

AM

 

[MechaMZ]

Since the object goes in a circle, the Tension + mg will always equal mv^2/R. The tension cannot go below 0 and mg cannot change, so the minimum force is mg. This means that the minimum circular speed is $\sqrt{gR}$. If v is less than this what happens at the top?

AM

thanks, so how about the second situation at B

http://img225.imageshack.us/img225/4176/p621.gif

 

[Andrew Mason]

thanks, so how about the second situation at B

http://img225.imageshack.us/img225/4176/p621.gif

At[/URL] the bottom, the forces are different. What additional force is there? At the top, what keeps the car moving in a circle? (hint: the net force is the sum of two forces acting in opposite directions). What happens if the force required to keep the car moving in a circular arc exceeds mg?

AM