Energy conservation problem (need closure)

[qswdefrg]

Homework Statement

the small mass m sliding without friction along the looped track is to remain on the track at all times, even at the very top of the loop of radius r.
a) calculate, in terms of the given quantities, the min. release height h.

if actual release height is 2h, calculate
b) normal force exerted by track at bottom of loop
c) normal force exerted by track at top of loop
d) normal force exerted by track after block exits loop onto flat section

Homework Equations

Ek = 0.5mv^2
Eg = mgh
Fc = mv^2/r

The Attempt at a Solution

a) total energy = mgh
mass must remain on track at top of loop, so

mgh = 0.5mv^2 + mg(2r)
h = 2.5r

b) Does the release height matter? Because if mass does not accelerate up nor down, then Fn =mg? Is this the same for d)?

c) Centripetal force keeps mass moving
Fn = mv^2/r - mg
Fn = m(v^2/r - mg)

I'm not sure if this is right either.

any pointers/hints would be greatly appreciated.

 

[tiny-tim]

hi qswdefrg!

(try using the X² icon just above the Reply box )

the mass will maintain contact with the track only if the https://www.physicsforums.com/library.php?do=view_item&itemid=73" is non-zero

find that reaction force by using F_total = ma, with a being the https://www.physicsforums.com/library.php?do=view_item&itemid=27"

start again! ​

 

[qswdefrg]

hi qswdefrg!

(try using the X² icon just above the Reply box )

the mass will maintain contact with the track only if the https://www.physicsforums.com/library.php?do=view_item&itemid=73" is non-zero

find that reaction force by using F_total = ma, with a being the https://www.physicsforums.com/library.php?do=view_item&itemid=27"

start again! ​

at the top of the loop

F_net = F_N - F_g = ma

F_N = mv²/r + mg

F_N = m(v²/r + g)

this is similar to the answer I had before. I realize now that I hadn't used the 2h at all. Am I supposed to use energy conservation to figure this out?

 

[tiny-tim]

hi qswdefrg!

yes, use conservation of energy to find what v is

(but your equation is wrong … a is in the same direction as g … anyway, how else would you get F_N = 0?)

 

[qswdefrg]

hi qswdefrg!

yes, use conservation of energy to find what v is

(but your equation is wrong … a is in the same direction as g … anyway, how else would you get F_N = 0?)

I got v²= g(2h-r)

So is the equation is F_n+F_g=ma?

Sub in velocity and I get F_n=2mg[(h/r)-1]. Is this correct?

 

[qswdefrg]

Never mind my previous post; I think I've got it now.

at the top:

v²= 4g(r-h)

F_net=F_c=F_n+F_g
F_n=mv²/r - mg
sub in v²
F_n= {[4mg(r-h)]/r } - mg

at the bottom:

v²= 4gh

F_net=F_c=F_n-F_g
F_n=mv²/r + mg
F_n= (4mgh/r) + mg

at the end:

F_n= mg

*fingers crossed*

 

[tiny-tim]

hi qswdefrg!

(just got up :zzz: …)

(b) and (c) look fine

(though you haven't finally converted h into 2.5r, and it would be easier to check if you put all the stages in)

but shouldn't this be 4g(h-r) ? …

v²= 4g(r-h)

 

[qswdefrg]

hi qswdefrg!

(just got up :zzz: …)

(b) and (c) look fine

(though you haven't finally converted h into 2.5r, and it would be easier to check if you put all the stages in)

but shouldn't this be 4g(h-r) ? …

Yes the h and r should be switched; silly error my bad.

finding h in part a):

mgh = 0.5mv² + mg2r
at the top min. speed is sqrt rg
mgh = 0.5mrg + mg2r
simplify
h = 2.5r

thanks so much for your help. i (finally!) got it.