Max Kinetic Energy of Object in Circular Path with Constraint

[tprofl]

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 tension exceeds 16N. The maximum kinetic energy the object can have is:

Homework Equations

KE=1/2mv^2
U=mgh

The Attempt at a Solution

Because the object is spinning in a horizontal circle, you may take the tension at any point. Potential can equal kinetic, mgh=1/2mv^2, but I can't figure out how to start this problem.

 

[cepheid]

Welcome to PF!

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 tension exceeds 16N. The maximum kinetic energy the object can have is:

Homework Equations

KE=1/2mv^2
U=mgh

The Attempt at a Solution

Because the object is spinning in a horizontal circle, you may take the tension at any point. Potential can equal kinetic, mgh=1/2mv^2, but I can't figure out how to start this problem.

I can't figure out what you're saying in your solution attempt, in particular, the bit about taking the tension "at any point."

Furthermore, since the circle is horizontal, the object never changes height, meaning that its gravitational potential energy is constant. Potential energy, therefore, is somewhat irrelevant to the problem.

Try, instead, to think of it this way: since kinetic energy depends on speed, the object's maximum kinetic energy is going to depend on its maximum possible speed. What determines the speed of an object in uniform circular motion? Hint: what kind of force is the tension in the rope providing in order to produce circular motion?

 

[PeterO]

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 tension exceeds 16N. The maximum kinetic energy the object can have is:

Homework Equations

KE=1/2mv^2
U=mgh

The Attempt at a Solution

Because the object is spinning in a horizontal circle, you may take the tension at any point. Potential can equal kinetic, mgh=1/2mv^2, but I can't figure out how to start this problem.

Because this object is on a horizontal table, its gravitational potential energy (mgh) doesn't ever change.

The centripetal force required to keep an object in circular motion is given by mv²/R

I was always interested in comparing the kinetic energy formula 1/2 mv² as they both contain the mv² expression

By comparing and combining those two expressions/formulas I think you can discover the answer you seek.