Recent content by (Ron)^2=-1

  1. (Ron)^2=-1

    Predicting Reality: The Role of Physics in Accurate Outcome Forecasts

    Welcome to PF Jake! And congrats on your accomplishments!
  2. (Ron)^2=-1

    Can a Teenager Specialize in Physics?

    Welcome to PF Nick!
  3. (Ron)^2=-1

    How Is Angular Momentum Calculated in Different Pivot Scenarios?

    Yes, you're right! D is the distance from the centre of mass axis (axis of symmetry). Always remember: The moment of inertia Icm is always with respect to the centre of mass, from which there is a symmetric distribution of mass (at least for homogeneous rigid bodies). So whenever you need to...
  4. (Ron)^2=-1

    Interested in Structural Engineering? Join Our Community Today!

    Welcome to Physics Forums James!
  5. (Ron)^2=-1

    How Is Angular Momentum Calculated in Different Pivot Scenarios?

    You need to find the total moment of inertia with respect to the new axis.
  6. (Ron)^2=-1

    How Is Angular Momentum Calculated in Different Pivot Scenarios?

    Yes it is. :smile:
  7. (Ron)^2=-1

    How Is Angular Momentum Calculated in Different Pivot Scenarios?

    I think what you're missing here is the parallel axis theorem.
  8. (Ron)^2=-1

    Question regarding block and pulley system with inertia

    BvU is right!
  9. (Ron)^2=-1

    Question regarding block and pulley system with inertia

    Everything looks correct. You're right, from the torques acting on the pulley you get: T2-T1=[(Mass of the pulley)*R*α]/2 = (I*α)/R and from the no slip condition you know that α=a/R therefore T2-T1=[(Mass of the pulley)*a]/2 = (I*a) / R^2 and solving for a, you get...
  10. (Ron)^2=-1

    Rolling without slipping over a plank

    Thank you so much!, I finally got the idea. So the centre of mass must, indeed, have acceleration in order for the pure rotation condition to hold.
  11. (Ron)^2=-1

    Rolling without slipping over a plank

    Sorry here I meant a reference frame fixed to the initial position
  12. (Ron)^2=-1

    Rolling without slipping over a plank

    What I thought was that because the plank can slide over the horizontal smooth surface and assuming Force < Max static friction, then the cylinder will start rotating but the centre of mass will not undergo translational motion and then the velocity will be = 0 (all with respect to a reference...
  13. (Ron)^2=-1

    Rolling without slipping over a plank

    Yes, you're right Mister. Although the problem says nothing about friction between the plank and the cylinder, I thought because of the condition of pure rotation there must be static friction, otherwise the cylinder would just slip.
  14. (Ron)^2=-1

    Rolling without slipping over a plank

    Here is the diagram, sorry I was having trouble with the link.
  15. (Ron)^2=-1

    Rolling without slipping over a plank

    https://lh3.googleusercontent.com/-FQFbA0n6chNSwQwUE9jUwaEgA_EDDnwEXjPbTQBRg=w1366-h768-rw Here is the diagram. If you can't see it here is the set-up The cylinder is on top of the plank, and the plank lies in a smooth horizontal frictionless surface. The cylinder has a radius R and also a...
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