Block in a verticle circle with friction

[GGDK]

Homework Statement

A block of mass 15g enters the bottom of a circular, vertical track with a radius R = 0.5m at an initial velocity of 4/ms. If the block loses contact with the track at an angle of  = 130, what is Wk, the work done by kinetic friction?
http://img94.imageshack.us/img94/1760/problem1t.png

Homework Equations

Work Energy theorem: (1/2)m(Vfinal)^2 - (1/2)m(Vinitial)^2
Work = Force * Displacement

The Attempt at a Solution

The 3 forces acting on the block: normal force, gravity, and friction.
http://img43.imageshack.us/img43/442/fbodydia.png
Before block loses contact,
Gravity has a y-component of .5(sin(40))
Normal force has a y-component of .5(sin(40)) and an x-component of .5(cos(40))

The initial kinetic energy is (1/2).015(4)^2 which equals .012 Joules.
Not really sure where to progress from here.

 

[PeterO]

Homework Statement

A block of mass 15g enters the bottom of a circular, vertical track with a radius R = 0.5m at an initial velocity of 4/ms. If the block loses contact with the track at an angle of  = 130, what is Wk, the work done by kinetic friction?
http://img94.imageshack.us/img94/1760/problem1t.png

Homework Equations

Work Energy theorem: (1/2)m(Vfinal)^2 - (1/2)m(Vinitial)^2
Work = Force * Displacement

The Attempt at a Solution

The 3 forces acting on the block: normal force, gravity, and friction.
http://img43.imageshack.us/img43/442/fbodydia.png
Before block loses contact,
Gravity has a y-component of .5(sin(40))
Normal force has a y-component of .5(sin(40)) and an x-component of .5(cos(40))

The initial kinetic energy is (1/2).015(4)^2 which equals .012 Joules.
Not really sure where to progress from here.

Since the block is about to leave the track there will be no Normal reaction force. It is the necessity of the reaction force that ensures the block remains in contact with the track up to that point.

Thus the component of the weight force towards the centre will be exactly the centripetal force requires.

Centripetal force is mv²/R , and since R is 0.5 this gives 2mv².

This just happens to be 4 times the kinetic energy at the point; so we can calculate the kinetic energy from the component of the weight force.

You can easily calculate the Kinetic energy at the bottom. There will be a certain amount of this transformed to potential energy, as it is higher up, and still the remaining kinetic energy above.
The "missing" energy is work done by kinetic friction.

 

[GGDK]

Thus the component of the weight force towards the centre will be exactly the centripetal force requires.

Centripetal force is mv²/R , and since R is 0.5 this gives 2mv².

This just happens to be 4 times the kinetic energy at the point; so we can calculate the kinetic energy from the component of the weight force.

You can easily calculate the Kinetic energy at the bottom. There will be a certain amount of this transformed to potential energy, as it is higher up, and still the remaining kinetic energy above.
The "missing" energy is work done by kinetic friction.

Thanks! So to double check would this be

mg = 9.8(.015) = 4KE
[9.8(.015)]/4 = (1/2)(0.015)(Vfinal)^2
4.9 = V^2
Vfinal = 2.21359m/s

(1/2)(0.015)(2.21359)^2 - (1/2)(0.015)(4)^2 = -.08325 Joules

 

[PeterO]

Thanks! So to double check would this be

mg = 9.8(.015) = 4KE
[9.8(.015)]/4 = (1/2)(0.015)(Vfinal)^2
4.9 = V^2
Vfinal = 2.21359m/s

(1/2)(0.015)(2.21359)^2 - (1/2)(0.015)(4)^2 = -.08325 Joules

First line is wrong: mg is the whole force of gravity, only the component of it towards the centre is used.

read my earlier post again - I said "component of the weight force"

 

[PeterO]

Thanks! So to double check would this be

mg = 9.8(.015) = 4KE
[9.8(.015)]/4 = (1/2)(0.015)(Vfinal)^2
4.9 = V^2
Vfinal = 2.21359m/s

(1/2)(0.015)(2.21359)^2 - (1/2)(0.015)(4)^2 = -.08325 Joules

Also you are not interested in the final speed; you only needed the final Kinetic Energy - which you re-calculated on the very next line

 

[galaticman]

Homework Statement

A block of mass 15g enters the bottom of a circular, vertical track with a radius R = 0.5m at an initial velocity of 4/ms. If the block loses contact with the track at an angle of  = 130, what is Wk, the work done by kinetic friction?
http://img94.imageshack.us/img94/1760/problem1t.png

Homework Equations

Work Energy theorem: (1/2)m(Vfinal)^2 - (1/2)m(Vinitial)^2
Work = Force * Displacement

The Attempt at a Solution

The 3 forces acting on the block: normal force, gravity, and friction.
http://img43.imageshack.us/img43/442/fbodydia.png
Before block loses contact,
Gravity has a y-component of .5(sin(40))
Normal force has a y-component of .5(sin(40)) and an x-component of .5(cos(40))

The initial kinetic energy is (1/2).015(4)^2 which equals .012 Joules.
Not really sure where to progress from here.

If you are doing the same problem that I am doing for PHYS 170 I am pretty sure that the radius is 0.3m not 0.5m

 

[GGDK]

So would the component of gravity just be .5*sin(40)?

 

[GGDK]

If you are doing the same problem that I am doing for PHYS 170 I am pretty sure that the radius is 0.3m not 0.5m

Yep, I changed the number on so I could try the problem again using different values.

 

[PeterO]

So would the component of gravity just be .5*sin(40)?

I saw this 0.5 earlier but ignored it as a typo.

Where are you getting 0.5 from?

EDIT: I think I see - you are finding a component of the Radius, and calling it a component of the weight .

 

[GGDK]

Oops didn't even realize that!

[9.8(0.015)sin(40)]/4 = KE
KE = .023622 Joules
Final - Initial = Total KE
.023622 - .12 = -.096378 Joules

 

[PeterO]

Oops didn't even realize that!

[9.8(0.015)sin(40)]/4 = KE
KE = .023622 Joules
Final - Initial = Total KE
.023622 - .12 = -.096378 Joules

Getting close: Don't forget a lot of that energy has not been "lost", merely converted to Potential energy because it is higher.

NOTE: even without friction, the block will fall off at some point if it wasn't going fast enough to begin with. Indeed if it is going too slow it won't even reach the 90 degree position, so just stop then slide back down the track.

 

[lele44]

why did you divide by velocity? (4)

 

[GGDK]

why did you divide by velocity? (4)

Centripetal force is mv2/R , and since R is 0.5 this gives 2mv2.

This just happens to be 4 times the kinetic energy at the point; so we can calculate the kinetic energy from the component of the weight force.

I didn't divide by velocity.

In the example problem centripetal force is (mv^2)/r with r as .5 so it would be equal to 2mv^2 which is basically 4 times kinetic energy because KE is (1/2)mv^2.

 

[PeterO]

I didn't divide by velocity.

In the example problem centripetal force is (mv^2)/r with r as .5 so it would be equal to 2mv^2 which is basically 4 times kinetic energy because KE is (1/2)mv^2.

Great response GGDK - it shows you understood what I was saying.

Just to let you know my original logic.

The block enters the loop with a certain amount of Kinetic Energy.
As it moves up the loop, some of that energy is converted to Potential energy, some is lost to friction, and some of it remains as kinetic energy.
The PE is easy(ish) to calculate, the trick is to calculate how much Kinetic energy remains.
The first thing I addressed in my first post was calculating that - using that convenient expression you quoted above.

NOTE:
I have always been intrigued by the similarity of the Kinetic Energy and Centripetal Force expressions; (1/2)mv² (or mv²/2) and mv²/R ; meaning that the numerical value of one can be simply transformed to the numerical value of the other.

This "connection" is useful for analysing a loop-the-loop in a roller coaster.
When a cart creeps over hill, plunges to the ground level then swings up through a loop, the kinetic energy at bottom and top of the loop can be calculated from change in PE. - so just mgΔh.

That expression gives mv²/2 , and the centripetal force resulting at each point is mv²/R - so it is easy to convert to the forces involved, without actually bothering with how fast the roller coaster is traveling at the time.