Recent content by MrDaahl

  1. M

    A small lump of ice sliding down a large, smooth sphere.

    The length OC must be 3/4 R and the length of AC must therefore be 1/4 R
  2. M

    A small lump of ice sliding down a large, smooth sphere.

    wait
  3. M

    A small lump of ice sliding down a large, smooth sphere.

    I know the R is not the same value in the initialising state as it is in the next. But i simly cannot see how to calculate PE_2
  4. M

    A small lump of ice sliding down a large, smooth sphere.

    KE1+PE1=KE2+PE2 KE1 = 0 PE1 = mgR KE2 = ½mv2 PE2 = mgΔR. → mgR=½mv2+mgΔR → this doesn't make sense :cry:
  5. M

    A small lump of ice sliding down a large, smooth sphere.

    That was wrong. PE1 = mgR
  6. M

    A small lump of ice sliding down a large, smooth sphere.

    Isnt it zero?
  7. M

    A small lump of ice sliding down a large, smooth sphere.

    Apparently not.. I assume I thought it should be the block's potentiel + kinetic energy at the initialising state, equal the kinetic energy at the secondary state.
  8. M

    A small lump of ice sliding down a large, smooth sphere.

    Ok, so I've to do something with R*2, which will equal the height of the circle So i got this equation so far: V1 = √(2*g*h+v02) Can anyone confirm the equation?
  9. M

    A small lump of ice sliding down a large, smooth sphere.

    Homework Statement A small lump of ice is sliding down a large, smooth sphere with a radius R. The lump is initially at rest. To get it started, it starts from a position slightly right to the sphere's top, but you can count it to start from the top. The lump is fallowing the sphere for a...