What is the Maximum Kinetic Energy of an Object Constrained by a Cord?

[hansel13]

Homework Statement

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?

Homework Equations

T=mg
KE=1/2mv²

The Attempt at a Solution

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

 

[teckid1991]

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

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

 

[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..

 

[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
(16*radius*mass)^.5 = v

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.

 

[hansel13]

(16*radius*mass)^.5 = v

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

 

[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