Recent content by johnschmidt

Homework Statement Solid disk, 50cm in diameter, with a mass of 5kg. Disk is located in space by a bearing. The friction torque of the bearing is 10Nm. A string is wrapped around the disk and connected to a weight. The weight is allowed to fall and spin the disk. How much energy is lost to...

Ok, I think I understand. This is like where energy is a function of velocity squared, right? Thanks haruspex!

Yes! Good catch. I guess one shouldn't complain about having too many equations ;-) Ok, then let's solve for T1 in terms of T2. ##\tau_{1} - \tau_{2} = I_{p1} \alpha_{p1} + \tau_{friction of p1}## ##T_{1} r_{p1} - T_{2} r_{p1} = \frac{m_{p1} r_{p1}^2}{2} \frac{a_{1}}{r_{p1}} +...

Homework Statement http://essbaum.com/images/Pulley_diagram.png We have a bonus question where we are asked to find the mass of ##m_{1}## necessary to accelerate ##m_{2}## upward at ##1g##, and we have to include friction in the pulley bearings as well as the pulley inertias. The...

I meant different TENSIONS, of course, not different torques. Thanks haruspex!

To be clear, the different torques are used when the pulley requires torque, but NOT simply because the pulley accelerates rotationally, correct?

Sometimes when I see pulley and rope questions people will say "for a non-extensible rope the tension is uniform throughout the rope". But I am now learning pulleys and rotational inertia and it seems to be necessary to consider the rope on either side of a pulley to have _different_ tension if...

So you are saying that the pressure necessary to support the cylinder is much higher than that provided by the weight of the fluid, yes? Hmm, Google suggests the pressure in the unopened can is 117 - 620kPa depending on temperature. Does anyone know of any equations related to supporting a...

I understand that a cylinder oriented vertically and bearing a load can be made more resistant to buckling by adding pressure to the (closed) cylinder. For example, an unopened can of soda pop can bear much more load than an opened soda pop can (I found a youtube video showing an unopened can...

Yes, you're right! The wrong sign on a was the key. Now I get: a1 = 2g(2m1 - m2) / (m1*p^2 + m2) Plugging in -9.8m/s^2 for g gives: m1 = 2 kg m2 = 2 kg a1 = -3.92 m/s^2 m1 = 1 kg m2 = 2 kg a1 = 0 m/s^2 m1 = 0.9 kg m2 = 2 kg a1 = 0.7 m/s^2 m1 = 0.5 kg m2 = 2 kg a1 =...

Hi barryj, You have the same problem I do where for certain values of m1 and m2 you get a zero in the denominator.

Homework Statement Note that the FAQ is not clear if I am required to use LaTeX for formatting formulas. I have not used LaTeX, but I will retype my question if I am violating the rules. Please let me know :-) Non-extensible cable attached to ceiling. Cable goes down, around a pulley and...