Question now has a picture to match

  • Thread starter Thread starter foxandthehen
  • Start date Start date
  • Tags Tags
    Match Picture
foxandthehen
Messages
13
Reaction score
0
http://img60.imageshack.us/img60/7782/question6jh6.png

The axis of a disk with radius of 0.05m is fitted into a slot that forms an angle of 30degs with the vertical axis. The disk is at rest when it is placed in contact with a convayor belt moving at contsant speed. The coefficent of kenetic friction between the disk and the belt is 0.2 and there is no bearing friction.

a) Express the acceleration of the center of mass of the body, and its angular acceleration

b) Draw a free body diagram showing the applie forces and the eqivalent force diagram showing the vectror 'ma' or its conponents and 'Ia'

c) Write the three equations of motion and use them to determine the angular acceleration of the disk while slipping occurs.

------------------------

HELP! Where should I start? Any help and advice would be very much appreciated.
 
Last edited by a moderator:
Physics news on Phys.org
Well, you're being asked for angular acceleration. What formula do you know for that?

That's right... tau = I * alpha, in other words, alpha = tau / I, where alpha is the angular acceleration, tau is the torque, and I is the moment of inertia. Does that get you started?
 
Hi! Thanks for the answer! I managed to do and hand it in yesterday and it was really quite simple in the end, think I just had a mental block about it! lol.
oh and I used Angular acceleration = (1/r)(Vbelt/dt) :D
 
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
##|\Psi|^2=\frac{1}{\sqrt{\pi b^2}}\exp(\frac{-(x-x_0)^2}{b^2}).## ##\braket{x}=\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dx\,x\exp(-\frac{(x-x_0)^2}{b^2}).## ##y=x-x_0 \quad x=y+x_0 \quad dy=dx.## The boundaries remain infinite, I believe. ##\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dy(y+x_0)\exp(\frac{-y^2}{b^2}).## ##\frac{2}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,y\exp(\frac{-y^2}{b^2})+\frac{2x_0}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,\exp(-\frac{y^2}{b^2}).## I then resolved the two...
Back
Top