Find the time for the string to break

[TheSwedeAtLarge]

Hi all!

I am currently working on an assignment for my dynamics course in which we need to create our own "real world" problems. I chose to do a problem which would solve the time at which the string of a lanyard would break when it is swung in a circle. My problem is with my end numbers. They seem very unreasonable but I am not sure if it is due to a calculation error, a dynamics error, or if it is simply the nature of the problem. I estimated the mass of the keys. The maximum load is from this website: http://www.engineeringtoolbox.com/polyester-rope-strength-d_1514.html
The torque of the motor is an arbitrary number that I chose.

Any feedback would be greatly appreciated.

1. Homework Statement

A motor is set up to spin a lanyard from rest with keys attached with a torque of 5 Nm. The polyester lanyard is 50 centimeters long with a diameter of 6 mm, assume it is massless. The keys have a mass m = 100 grams. The maximum load of polyester rope is Tmax = 3400 N. At what time will the rod break and after how many revolutions? Gravity acts in the -j direction. Neglect bending stresses and air resistance.

Schematic: http://imgur.com/V6mSuuD
FBD/KD: http://imgur.com/AQShsEv

Homework Equations

F=ma
w = w0 + at

The Attempt at a Solution

Find Angular Velocity[/B]
Sum forces in the y-direction and solve for w

Tmax – mg = mrw2
w = 260.7 rad/s

Find Angular Acceleration
Sum moments about the origin

M = (mra) * (r)
a=M/mr^2

a = 200 rad/s2

Use kinematics to find the time to break

w = w0 + at

260.7 rad/s = 200 rad/s2 * t

t = 1.3 s

 

[TSny]

Welcome to PF!

I think your calculation is OK. It just shows that 5 Nm is a "large" torque in this situation.

If you thought of the torque as due to a force F applied tangentially to the circular motion of the keys, then F would be 10 N. That might not sound like much, but that force would give the keys an acceleration of F/m = 100 m/s². Thus, starting from rest, this acceleration would cause the keys to travel 200 meters in 2 seconds.

Did you calculate the number of revolutions of the lanyard in 1.3 s?

 

[TheSwedeAtLarge]

Thanks for the feedback! I was confused on what to set the torque at for it to be more realistic but I don't know what would work.

I think it gives 53.96 revolutions. Which is a ton in 1.3 seconds!

 

[TSny]

Thanks for the feedback! I was confused on what to set the torque at for it to be more realistic but I don't know what would work.

I think it gives 53.96 revolutions. Which is a ton in 1.3 seconds!

I think it's about half that number of revolutions. Still a lot.