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M2 Normal reaction question Watch

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    Does (i) mean calculate the normal reaction force? If so why does the mark scheme say to use Rcos(a)=mg rather than mgcos(a)=R?
    Isn't mgcos(a)=R now you usually find the normal reaction force on a slope?
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    (Original post by 000alex)


    Does (i) mean calculate the normal reaction force? If so why does the mark scheme say to use Rcos(a)=mg rather than mgcos(a)=R?
    Isn't mgcos(a)=R now you usually find the normal reaction force on a slope?
    Since the particle is moving in a horizontal circle, which way is the acceleration?

    So which way do the forces cancel out?
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    (Original post by 000alex)


    Does (i) mean calculate the normal reaction force?
    Yes, they mean the normal reaction force. Since the surface is smooth, the reaction is purely normal.

    If so why does the mark scheme say to use Rcos(a)=mg rather than mgcos(a)=R?
    Isn't mgcos(a)=R now you usually find the normal reaction force on a slope?
    Good question. The answer isn't at all clear to many people who are learning mechanics. They say: "Hmm, there is no acceleration in the direction of the line defined by R, since the particle does not leave the surface, hence resolving along the normal to the plane, we have R=mg\cos \theta"

    Sadly, this is incorrect, and they will get the answer wrong. Why? Well, it is *given* in the question that the particle is executing horizontal circular motion. Hence, you can draw a horizontal acceleration vector on P pointing toward the centre of the circle. This vector can be resolved into a component parallel to R and one perp to R.

    By Newton II, there is thus a net force parallel to R, and so R-mg\cos \theta \ne 0 \Rightarrow R \ne mg \cos \theta
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    (Original post by atsruser)
    y
    (Original post by tiny hobbit)
    Since the particle is moving in a horizontal circle, which way is the acceleration?

    So which way do the forces cancel out?
    Ok I get why R can't be mgcos(a).

    Can someone help with the math that shows why Rcos(a)=mg? I can't get anything to cancel out
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    (Original post by 000alex)
    Ok I get why R can't be mgcos(a).

    Can someone help with the math that shows why Rcos(a)=mg? I can't get anything to cancel out
    The only acceleration is horizontal, so upward force = downwards force.
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    Ok not really needed for this question but how would you work out the total force acting up the plane/slope?
    And would this force be equal to mgSin(a) since the particle isn't moving up or down the slope?
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    (Original post by 000alex)
    Ok not really needed for this question but how would you work out the total force acting up the plane/slope?
    And would this force be equal to mgSin(a) since the particle isn't moving up or down the slope?

    Isn't that the reaction force since the particle is in contact with the plane?
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    (Original post by MsFahima)
    Isn't that the reaction force since the particle is in contact with the plane?
    I thought it couldn't be because the reaction is perpendicular to the surface so there is no component of it up or down the plane
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    (Original post by 000alex)
    I thought it couldn't be because the reaction is perpendicular to the surface so there is no component of it up or down the plane
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    Hope this helps

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    (Original post by MsFahima)
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    Hope this helps

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    I get the R part of the question

    I'm just wondering what is stopping the particle from moving down the slope?



    What would ? be provided by?
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    (Original post by 000alex)
    I get the R part of the question

    I'm just wondering what is stopping the particle from moving down the slope?



    What would ? be provided by?
    Why do you need to know that? There is no frictional force since it's a smooth plane. Also, since it's moving at a constant angular speed, there is a centripetal force which holds it in place. Right? Correct me if I'm wrong!
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    (Original post by MsFahima)
    Why do you need to know that? There is no frictional force since it's a smooth plane. Also, since it's moving at a constant angular speed, there is a centripetal force which holds it in place. Right? Correct me if I'm wrong!
    I don't need it for this question, I'm just interested in what happens in case a hard question comes up in the exam.

    Doesn't the component of the centripetal force act down the plane not up it because the centripetal force is towards the center of the circle??
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    (Original post by 000alex)
    I don't need it for this question, I'm just interested in what happens in case a hard question comes up in the exam.

    Doesn't the component of the centripetal force act down the plane not up it because the centripetal force is towards the center of the circle??
    In the exam, if there is a frictional force they would say "rough plane" or tell you to work out the frictional force so you don't have to worry! No? It's towards the centre... so horizontally.
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    (Original post by MsFahima)
    In the exam, if there is a frictional force they would say "rough plane" or tell you to work out the frictional force so you don't have to worry! No? It's towards the centre... so horizontally.
    Yea but if the plane is smooth something must be keeping it from slipping, and aren't the components of F (Centripetal force) like this?
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    (Original post by MsFahima)
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    Hope this helps

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    This isn't correct. There is a component of acceleration along the line that you have chosen to resolve.
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    (Original post by 000alex)
    Yea but if the plane is smooth something must be keeping it from slipping, and aren't the components of F (Centripetal force) like this?
    No. Your F is not normal.

    There is a component of mg tangent to the slope.
    There is a component of mg normal to the slope.

    These forces, together with the normal reaction, cause accelerations tangent and normal to the slope. The resultant of those accelerations (which describes the true motion of the particle) points towards the centre of the circle. This is the centripetal acceleration.

    You can't analyse the complete motion of the particle by ignoring some of the accelerations that are acting on it.
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    (Original post by atsruser)
    This isn't correct. There is a component of acceleration along the line that you have chosen to resolve.
    No.. the acceleration is horizontal.
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    (Original post by atsruser)
    No. Your F is not normal.

    There is a component of mg tangent to the slope.
    There is a component of mg normal to the slope.

    These forces, together with the normal reaction, cause accelerations tangent and normal to the slope. The resultant of those accelerations (which describes the true motion of the particle) points towards the centre of the circle. This is the centripetal acceleration.

    You can't analyse the complete motion of the particle by ignoring some of the accelerations that are acting on it.
    So what counters the component of mg tangential to the slope? (so that the particle does not slip)

    Is it the component of the centripetal acceleration?
    I tried to draw the force due to this centripetal acceleration on the diagram but the tangential component of it points downwards
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    (Original post by MsFahima)
    Also, since it's moving at a constant angular speed, there is a centripetal force which holds it in place. Right? Correct me if I'm wrong!
    This suggests to me that you are not clear on the mechanism by which centripetal force causes uniform circular motion.

    The centripetal force causes an acceleration of the particle towards the centre of the circle. Over a short period of time, this acceleration creates a small velocity vector pointing toward the centre of the circle. The resultant of this centripetal velocity and the existing tangential velocity creates a new tangential velocity pointing in a slightly different direction, but of the same magnitude.

    It can't be said that the centripetal force holds the particle in position though.
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    (Original post by MsFahima)
    No.. the acceleration is horizontal.
    Please see my post above for a more complete explanation. The horizontal acceleration vector a has a component of a \cos \theta in the direction of the normal reaction force. By Newton II, there must be a force creating this acceleration, which is R-mg\cos\theta.

    You are correct in saying that the acceleration is horizontal, however. This means that the *only* direction in which it has no component is at 90 degrees to itself i.e. vertically. Hence, there is no nett *vertical* force giving R\cos\theta=mg
 
 
 
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