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ntukza

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About ntukza

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  1. If they are opposite then surely they can't both point in the same direction, right? Merged post follows: Consecutive posts mergedI've figured it out guys, thanks for your help. The only thing I don't fully understand is why weight = centripetal force if centripetal force always acts toward the centre of the circle. But assuming it does: mg=m(v^2)/r v^2 = g.r = 80 Now the energy equations: mgh = mg(2a) + 1/2m(v^2) for 2a=16 and (v^2)=80, h=20 Once again thanks for helping me see that I needed to consider centripetal force
  2. Ok, it's starting to make sense now. So the centripetal force needs to act away from the centre of the circle so as to balance the weight which acts towards the centre of the circle? mg = m(v^2)/r v^2 = g.r So we can find v at the top, and then what do we do from there?
  3. I didn't think of that, thanks. But I'm still a bit confused. How do I take the centripetal force into consideration? If the object leaves the surface of the sphere eventually than surely the centripetal fore, which should act towards the centre of the circle, is not the net force... N = mgcosL + m(v^2)/r ?
  4. The way I see it, if the roller coaster starts at 16m above the ground, its mechanical energy is mgh and obviously v=0. It will therefore reach the top of the vertical loop of diameter 16m but will not go past the top point because there v=0 and E=mgh again, so it will fall off. For it to remain on the tracks h must be greater than 16. This means even h=17m will be sufficient. Why does h have to be 20m. Can someone please explain to me
  5. A bead slides down the surface of a sphere. Ignoring the initial speed of the bead, define the normal force between the bead and the sphere with respect to the angle of displacement. Given Answer: N = mg(3cosL-2) , L is the angle between the bead and the vertical taken from the centre of the sphere. My answer was N = mgcosL. I don't see how the given answer could have been derived.
  6. A bead slides down the surface of a sphere. Ignoring the initial speed of the bead, define the normal force between the bead and the sphere with respect to the angle of displacement. Given Answer: N = mg(3cosL-2) , L is the angle between the bead and the vertical taken from the centre of the sphere. My answer was N = mgcosL. I don't see how the given answer could have been derived.
  7. A roller coaster freewheels down a frictionless track from rest a height h above the ground. At the bottom the track becomes a vertical loop with diameter 2a = 16m. How high above the ground (value of h) must the roller coaster start if it is to reach the top of the vertical loop without lifting off the track? The given answer is 20m. I would have thought that the roller coaster needs to start at a height equal to, or slightly more than, the diameter of the vertical circle at the bottom. Energy at start = energy at end, therefore mgh = mg2a, and h = 2a = 16m. Please help me understand why
  8. A roller coaster freewheels down a frictionless track from rest a height h above the ground. At the bottom the track becomes a vertical loop with diameter 2a = 16m. How high above the ground (value of h) must the roller coaster start if it is to reach the top of the vertical loop without lifting off the track? The given answer is 20m. I would have thought that the roller coaster needs to start at a height equal to, or slightly more than, the diameter of the vertical circle at the bottom. Energy at start = energy at end, therefore mgh = mg2a, and h = 2a = 16m. Please help me understand why
  9. //"The wording of the question is a bit ambiguous, but I'm sure the intended meaning is that the trolley isn't moving before the throw."// Yes you're right. I didn't realise that the trolley could actually be motionless initially. It makes sense now that you've mentioned it. Thanks a lot! //"That equation is incorrect. You’re using kinetic energy instead of momentum, which is just mv. The total kinetic energy is not going to be the same before and after, since you're using potential energy in your arm to throw the rock!"// Yeah I was kind of worried about that. Thanks for cla
  10. Momentum is conserved (the trolley will have the same momentum in the opposite direction as the rock) But what about the initial momentum that also needs to be conserved? Surely your statement applies if there was no momentum to begin with, because the momentum before the throw must equal the momentum after the throw - The sum of the trolley's and stone's momentum must equal the momentum before the throw, ratheer than them equalling each other... Having said that I think you've helped me understand a little better. Would it be correct to say: 1/2(M+m)u^2 = [1/2(M+m)u^2 - 1/2M(J/M
  11. 1/2(M+m)v^2 = 1/2Mv^2 + 1/2m(J/m)^2 Mv^2 + mv^2 - Mv^2 = m(J/m)^2 mv^2 = m(J/m)^2 v^2 = (J/m)^2 v = J/m This assumes v is the same before and after collision, an assumption that I don't fully agree with. Without this assumption the equation yields another variable, which then becomes hard to solve. Maybe the given answer of J/M is wrong. Merged post follows: Consecutive posts merged Wouldn't the trolley still have Kinetic Energy since it was said to be moving originally?
  12. Please help me with this question: A man sits on a trolley, which runs without friction on horizontal rails. The mass of the trolley (with the man included) is M. The man throws horizontally a stone of mass m, in the direction of the rails. Find the subsequent speed of the trolley if: (i) The momentum imparted by the man to the stone is J; (ii) The total work done by the man is W. The answers given are (i) v = J/M and (ii) v = squareroot[2mW/M(M+m)] For the first part I got J/m and I assumed the trolley's speed did not change after the stone was thrown, which doesn't make sense to
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