Difference between revisions of "User:Msf24"

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let <math>I</math> be the rotational inertia of the roll  <math>I</math>= <math>\frac{1}{2}(M)(R)^2</math>
 
let <math>I</math> be the rotational inertia of the roll  <math>I</math>= <math>\frac{1}{2}(M)(R)^2</math>
  
And let <math>\tao</math>be the torque
+
let <math>R</math> be the radius of the roll of toilet paper
 +
 
 +
And let <math>\tau</math>be the torque
 +
 
 +
now in order to rip the toilet paper, <math>\tau = f \times R</math>
  
  

Revision as of 03:42, 30 December 2009

Work to see how fast one must pull on toilet paper to get it to tear:

let \(f\) be the force required to tear it

let \(M\) be the Mass of the toilet paper roll

let \(I\) be the rotational inertia of the roll \(I\)= \(\frac{1}{2}(M)(R)^2\)

let \(R\) be the radius of the roll of toilet paper

And let \(\tau\)be the torque

now in order to rip the toilet paper, \(\tau = f \times R\)





Variations/Modifications:

You could require the students to derive \(I\)

You could require the students to calculate \(M\) given information about the dimensions and how heavy some toilet paper is