thank you so much for these videos they are extremely helpful! would you be able to explain how an approach looking at centripetal force would work for the last question?
Thank you, I'm glad they are useful. I don't know how clear this explanation will be squeezed in a comment box but I'll give it a go. Let the circular part of the track have radius R, and the slider (mass M) be at an angle A above the horizontal radius. The kinetic energy of the slider will be Mg(R - RsinA) from conservation of energy. If the slider is moving in circular motion along the track then it must experience a centripetal force of Mv^2/R, which is the kinetic energy multiplied by 2/R, so 2Mg(1 - sinA). Resolving forces (on the slider) radially gives the normal reaction force, N, plus the component of the weight, MgsinA. Furthermore, N must be positive else the slider isn't touching the track. So that leaves us with N + MgsinA=2Mg(1 - sinA) where N is positive (or zero at the point the slider leaves the track). Rearranging gives N=2Mg-3MgsinA for non-negative N, so if sinA>2/3 the slider must have left the track. This gives an upper limit to how far the slider can move along the track. You might need to write this out for it to make sense!
@@BlakeTerrorPhysics Hi, I've been struggling with this problem for a while and I understood this explanation. I like how it's also very thorough and gives a precise value for the position at which the slider will break contact with the plane. Still, I want to understand why couldn't I say directly that not all the GPE from the beginning is converted into linear kinetic energy because the slider also rotates. Part of the kinetic energy will be rotational kinetic energy. I have a feeling that this aproach does not work because the slider is not spining in respect to an axis that goes thorugh it. There's just some ambiguity around the kinetic energy and when can you talk about rotational kinetic energy and I was wondering if you could've cleared that out for me. Thank you so much, keep up the great work!
@@robertlascar6695 A few points on the rotation of the slider that might help. Firstly, you're not wrong to think about rotational energy - the slider is rotating, and will continue to rotate when it leaves the track (each end of the slider will have a different velocity when it leaves the track so it will keep on spinning). Secondly, it is more difficult to analyse the rotational motion than the linear motion. Thirdly, in the context of this question, your intuition about rotational energy doesn't seem to extend far enough to rule out answer B (i.e. if the slider reached point T and stopped, then it wouldn't have either linear or rotational KE, so how do we know that it couldn't have got there?). So what do we do about all this? This nicely uncovers some unstated assumptions in the circular motion maths that I included in my previous comment: the slider will reach its highest point on the track when all KE is linear, therefore if the slider is treated as a point mass and it still can't reach point T, then it also won't reach point T if the slider has length. So the sinA=2/3 limit is an upper limit for the slider modelled as a point mass. If the slider has length then we would need to do the (rather more difficult) maths again.
thank you so much for these videos they are extremely helpful! would you be able to explain how an approach looking at centripetal force would work for the last question?
Thank you, I'm glad they are useful. I don't know how clear this explanation will be squeezed in a comment box but I'll give it a go. Let the circular part of the track have radius R, and the slider (mass M) be at an angle A above the horizontal radius. The kinetic energy of the slider will be Mg(R - RsinA) from conservation of energy. If the slider is moving in circular motion along the track then it must experience a centripetal force of Mv^2/R, which is the kinetic energy multiplied by 2/R, so 2Mg(1 - sinA). Resolving forces (on the slider) radially gives the normal reaction force, N, plus the component of the weight, MgsinA. Furthermore, N must be positive else the slider isn't touching the track. So that leaves us with N + MgsinA=2Mg(1 - sinA) where N is positive (or zero at the point the slider leaves the track). Rearranging gives N=2Mg-3MgsinA for non-negative N, so if sinA>2/3 the slider must have left the track. This gives an upper limit to how far the slider can move along the track. You might need to write this out for it to make sense!
@@BlakeTerrorPhysics Hi, I've been struggling with this problem for a while and I understood this explanation. I like how it's also very thorough and gives a precise value for the position at which the slider will break contact with the plane. Still, I want to understand why couldn't I say directly that not all the GPE from the beginning is converted into linear kinetic energy because the slider also rotates. Part of the kinetic energy will be rotational kinetic energy. I have a feeling that this aproach does not work because the slider is not spining in respect to an axis that goes thorugh it. There's just some ambiguity around the kinetic energy and when can you talk about rotational kinetic energy and I was wondering if you could've cleared that out for me. Thank you so much, keep up the great work!
@@robertlascar6695 A few points on the rotation of the slider that might help. Firstly, you're not wrong to think about rotational energy - the slider is rotating, and will continue to rotate when it leaves the track (each end of the slider will have a different velocity when it leaves the track so it will keep on spinning). Secondly, it is more difficult to analyse the rotational motion than the linear motion. Thirdly, in the context of this question, your intuition about rotational energy doesn't seem to extend far enough to rule out answer B (i.e. if the slider reached point T and stopped, then it wouldn't have either linear or rotational KE, so how do we know that it couldn't have got there?). So what do we do about all this? This nicely uncovers some unstated assumptions in the circular motion maths that I included in my previous comment: the slider will reach its highest point on the track when all KE is linear, therefore if the slider is treated as a point mass and it still can't reach point T, then it also won't reach point T if the slider has length. So the sinA=2/3 limit is an upper limit for the slider modelled as a point mass. If the slider has length then we would need to do the (rather more difficult) maths again.