Difference between revisions of "User:Msf24"
Jump to navigation
Jump to search
| Line 5: | Line 5: | ||
let <math>M</math> be the Mass of the toilet paper roll | let <math>M</math> be the Mass of the toilet paper roll | ||
| − | 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> because it is a solid cylinder |
let <math>R</math> be the radius of the roll of toilet paper | let <math>R</math> be the radius of the roll of toilet paper | ||
| Line 11: | Line 11: | ||
And let <math>\tau</math>be the torque | And let <math>\tau</math>be the torque | ||
| − | now in order to rip the toilet paper, <math>\tau = f \times R</math>, and we also know that <math>\tau = \frac{1}{2} ( | + | now in order to rip the toilet paper, <math>\tau = f \times R</math>, and we also know that <math>\tau = \frac{1}{2} (I)(\omega)^2</math> |
Revision as of 03:44, 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 \(I\)= \(\frac{1}{2}(M)(R)^2\) because it is a solid cylinder
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, \(\tau = f \times R\), and we also know that \(\tau = \frac{1}{2} (I)(\omega)^2\)
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
