Hey there! Sign in to join this conversationNew here? Join for free
x Turn on thread page Beta
    • Thread Starter
    Offline

    3
    ReputationRep:
    Hey,

    Question: https://08c6cd28b8bef288858e878e5745...l%20Fields.pdf

    see page 13 q3b(i)

    i understand that the area underneath a gravitational force-distance graph = potential energy, and so you could just count the squares to work this out. I also understand that you could find the potential energy by calculation: i.e. change in potential = potential at surface - 0, then multiply this value by 1000kg to get the change in potential energy.

    But my question is why can't you find the kinetic energy of the 1000kg mass at the surface of the moon, i.e. 0.5m(squared), where v is the speed at the surface found from equating the centripetal force to the gravitational force? Surely this 'gain' in ke is equivalent to the loss in pe? Or can I not work out v like that because the mass is not in 'orbit' around the surface (which is what equating the centripetal force to the gravitational force implies)

    Thanks!

    uberteknik
    Offline

    2
    ReputationRep:
    (Original post by Nikhilm)
    Hey,.....
    Or can I not work out v like that because the mass is not in 'orbit' around the surface (which is what equating the centripetal force to the gravitational force implies)
    uberteknik
    That's why net force is not equal to centripetal force in this case. You are right. You can't work this way.

    (Original post by Nikhilm)
    Hey,.....
    this 'gain' in ke is equivalent to the loss in pe?uberteknik
    Certainly throwing vertically upwards any object the gain in P.E. is equal to the loss in K.E. As soon as the object reaches a point which is at a very large distance from Moon's surface the final P.E. is almost zero. Moreover, how much K.E. is left when the mass reaches that point ?
    Offline

    20
    ReputationRep:
    It is equal, but you have no way of working out the change in kinetic energy without first working out the change in potential energy.
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by morgan8002)
    It is equal, but you have no way of working out the change in kinetic energy without first working out the change in potential energy.
    So the value of v in the centripetal force = gravitational force equation can't be used in the change in KE equation for the reason I stated in my origin post?
    Offline

    20
    ReputationRep:
    (Original post by Nikhilm)
    So the value of v in the centripetal force = gravitational force equation can't be used in the change in KE equation for the reason I stated in my origin post?
    You're correct that centripetal force only makes sense if the object is in some kind of orbit and we don't know whether or not it is. It's just a semantic problem though because net force can be used where centripetal force doesn't make sense.

    The value of v could be found in that way I think(if you calculate the cases where the object is not in orbit separately), but you'd have to derive the centripetal force in terms of v for an elliptical orbit. It seems my previous post was incorrect. Unfortunately I don't know enough differential geometry yet to calculate it personally. In any case it is a lot more advanced than the maths that would be expected in an A-level physics exam. Physics is all about finding the easiest way to solve a problem and in this case conservation of energy is the easiest.
    To be honest I'm not sure if that makes sense. I'll look at it tomorrow.
    Offline

    2
    ReputationRep:
    It is not so complicated. According to my experience, the way the question is set helps you think about your next steps. You asked to find the escaping velocity and the previous question asks about the change in Potential Energy. You could think that you gain potential energy as you gain distance from the moon, going towards to the (theoretically) infinite distance from moon's surface. You lose exactly the same amount of potential energy when you travel from a point which is theoretically at an infinite distance from the moon to a point which is at the moon's surface. You can write down your reasoning to explain this.

    (Original post by Nikhilm)
    So the value of v in the centripetal force = gravitational force equation can't be used in the change in KE equation for the reason I stated in my origin post?
    (Original post by morgan8002)
    Y In any case it is a lot more advanced than the maths that would be expected in an A-level physics exam. Physics is all about finding the easiest way to solve a problem and in this case conservation of energy is the easiest.
    To be honest I'm not sure if that makes sense. I'll look at it tomorrow.
    Offline

    20
    ReputationRep:
    (Original post by depymak)
    It is not so complicated. According to my experience, the way the question is set helps you think about your next steps. You asked to find the escaping velocity and the previous question asks about the change in Potential Energy. You could think that you gain potential energy as you gain distance from the moon, going towards to the (theoretically) infinite distance from moon's surface. You lose exactly the same amount of potential energy when you travel from a point which is theoretically at an infinite distance from the moon to a point which is at the moon's surface. You can write down your reasoning to explain this.
    Yeah, I was just saying that I was wrong before and there is another way of working it out without reference to energy, except in the final application of conservation of energy. Using energy directly is the fastest and simplest way by far and should obviously be done in this question.
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by morgan8002)
    You're correct that centripetal force only makes sense if the object is in some kind of orbit and we don't know whether or not it is. It's just a semantic problem though because net force can be used where centripetal force doesn't make sense.

    The value of v could be found in that way I think(if you calculate the cases where the object is not in orbit separately), but you'd have to derive the centripetal force in terms of v for an elliptical orbit. It seems my previous post was incorrect. Unfortunately I don't know enough differential geometry yet to calculate it personally. In any case it is a lot more advanced than the maths that would be expected in an A-level physics exam. Physics is all about finding the easiest way to solve a problem and in this case conservation of energy is the easiest.
    To be honest I'm not sure if that makes sense. I'll look at it tomorrow.
    Hey, really quick question about EMI.

    For magnetic flux, BA, what does A specifically mean? Does it mean the area swept out by the conductor if it's moving relative to the magnetic field, or just the area of the conductor itself, or something else?
    Offline

    20
    ReputationRep:
    (Original post by Nikhilm)
    Hey, really quick question about EMI.

    For magnetic flux, BA, what does A specifically mean? Does it mean the area swept out by the conductor if it's moving relative to the magnetic field, or just the area of the conductor itself, or something else?
    If \epsilon is the emf around a loop, A is the area of the flat surface with its edges forming the loop. This only works if B is always normal to the surface and constant across the surface(but may depend on time). The real definition of flux is much more general.
    You also need to know the case where \vec{B} is at an angle \theta to the normal of the surface, in which case \Phi_B = BA\cos \theta.

    Eg. if you have a circular loop and a field that is constant and perpendicular to the plane of the circle, then the surface that you'd use is the area inside the circle.

    For conductor that is a single straight wire, we'd use a rectangular loop with one of the edges being the side of the conductor(the other three don't matter as long as they're there). The side opposite the conductor is fixed and conductor side moves with the conductor. The loop will get bigger, so the flux will increase.
    For a loop conductor we'd have to use the conductor itself as the loop since we want the emf of the loop itself. The surface would be the surface inside of the loop.
    Offline

    2
    ReputationRep:
    Name:  magnetic flux.JPG
Views: 62
Size:  34.1 KB
    generally,A represents any surface. In picture 1 -3 , A is the area defined by the part of the circuit which is inside the magnetic field.
    In picture 1, magnetic flux changes because the area changes orientation with respect to magnetic field lines.
    In picture 2 magnetic flux changes because the magnetic field changes due to the movement of the magnet
    In picture 3 the area itself changes. In this case in order to caluclate the change in magnetic flux you need to consider the area swept out by the moving part of the circuit
    (Original post by Nikhilm)
    Hey, really quick question about EMI.

    For magnetic flux, BA, what does A specifically mean? Does it mean the area swept out by the conductor if it's moving relative to the magnetic field, or just the area of the conductor itself, or something else?
    • Thread Starter
    Offline

    3
    ReputationRep:
    depymak morgan8002 thank you so much for your help!
 
 
 
Poll
Are you going to a festival?

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.