Homework Help: Maximum kinetic energy

1. Dec 7, 2008

hansel13

1. The problem statement, all variables and given/known data
An object is constrained by a cord to move in a circular path of radius 0.5m on a horizontal frictionless surface. The cord will break if its tension exceeds 16 N. What is the maximum kinetic energy?

2. Relevant equations
T=mg
KE=1/2mv2

3. The attempt at a solution

Really lost on this problem. Am I applying the wrong formulas?

2. Dec 7, 2008

teckid1991

if its moving in a circular horizontal path the tension isn't mg its equal to the centripetal force (mv2/r)

so im guessing you set the centripetal force to equal 16 to solve for v. After that you can find the KE

3. Dec 7, 2008

hansel13

thanks teckid.

So a = v2/R

so V = (a*R)1/2
V = (9.8*.5)1/2
V = 2.21

T=(mv2)/R
m = T*R/(v2)
m = 1.63

So for KE, we have:
KE = 1/2*m*v2
KE = 1/2*1.63*2.212
KE = 4 J

I think that's right..

4. Dec 7, 2008

teckid1991

No i think you're confusing horizontal and vertical... vertical suggest there is a downward force due to gravity (aka 9.81...). However, here the path is horizontal, like a merry-go-round, so there gravity does not effect the tension on a frictionless surface. since the tension must be 16 or less and the tension is equal to the centripetal force then:

16 = mv2/r

further...

16 *radius * mass = v2

with this velocity you can find the max KE (.5MV2)

You should not be using centripetal acceleration to find the velocity unless we know what that acceleration is.

5. Dec 8, 2008

hansel13

But this leaves us with 2 unknown variables (mass and velocity)...

6. Dec 8, 2008

LowlyPion

Not really.

Maximum F = mv²/r = 16N

But Kinetic Energy is ½mv²

But you know that mv² = 16*r

Which means your KE is merely ½*16*r