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    Guys, it's \sqrt{5gl} since we have u^2=v^2+4gl and we require T \geq 0 at the top of the circle, hence \Sigma F=\dfrac{mv^2}{l}=mg+T \Rightarrow \dfrac{mv^2}{l} \geq mg \Rightarrow v^2 \geq gl \Rightarrow u^2=v^2+4gl \geq gl+4gl=5gl \Rightarrow u \geq \sqrt{5gl}.
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    (Original post by Jm098)
    Did anyone get 2 rootlg?

    I did for 5d
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    How can so many people not see that it's just 2 root lg? I'm confused
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    (Original post by aoxa)
    No, tension was root(4gl) which simplified, was 2root(gl). Which gives non negative tension.


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    i think you're forgetting about the gravitational force of the weight


    at top (mv^2)/r = T + mg

    for complete rotation T>>0 so

    (mv^2)/r = mg

    this gave v= root(5gl)
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    (Original post by jjsnyder)
    Here are my answers from todays exam, not all of them will necessarily be correct so please correct me if you think they are wrong!
    1. a) 9.6
    b) i) 24.3
    ii) 12.7
    iii) No Air Resistance
    2.
    a)  a = (8-4x^3)i - 18e^{-3x}j
    b) i)  F = (16-8x^3)i - 36e^{-3x}j
    ii) 8.20
    c)  r = (4x^2- \frac{x^5}{5} + 3)i - (3+ 2e^{-3x})j
    3. a) By symmetry
    b)10.4 cm
    c) 16.5 degrees
    d) Explanation Q
    4. a)78.4
    b) 41.4 degrees
    c) r=2.89
    5. Lots of very long answers, will upload later if I have time. Part d) was  sqrt(5gl/2)
    6. a) Show that Q
    b)  \frac{g-(g-\lambda\mu)e^{-t\lambda} }{\lambda}
    7. a) Diagram
    b)  tanx = \frac{1-2\mu^2}{4\mu}
    8. a) 17.0
    b) I got 6.05m but many others got 7.58m so this seems like the correct answer, can anyone confirm this?
    Question 5) was just root 5gl
    Question 8) I got 6.87 I think but forgot to include the potential energy still in the bottom string I think!
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    (Original post by IrrationalRoot)
    Guys, it's \sqrt{5gl} since we have u^2=v^2+4gl and we require T \geq 0 at the top of the circle, hence \Sigma F=\dfrac{mv^2}{l}=mg+T \Rightarrow \dfrac{mv^2}{l} \geq mg \Rightarrow v^2 \geq gl \Rightarrow u^2=v^2+4gl \geq gl+4gl=5gl \Rightarrow u \geq \sqrt{5gl}.
    Mate this exact question has come up before word for word and the answer is 2rootlg
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    (Original post by Jm098)
    How can so many people not see that it's just 2 root lg? I'm confused
    Because it's not.
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    (Original post by CoIIins)
    Mate this exact question has come up before word for word and the answer is 2rootlg
    Did you even read my method...
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    (Original post by IrrationalRoot)
    Guys, it's \sqrt{5gl} since we have u^2=v^2+4gl and we require T \geq 0 at the top of the circle, hence \Sigma F=\dfrac{mv^2}{l}=mg+T \Rightarrow \dfrac{mv^2}{l} \geq mg \Rightarrow v^2 \geq gl \Rightarrow u^2=v^2+4gl \geq gl+4gl=5gl \Rightarrow u \geq \sqrt{5gl}.
    Yes that is right you the mvp
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    (Original post by IrrationalRoot)
    Guys, it's \sqrt{5gl} since we have u^2=v^2+4gl and we require T \geq 0 at the top of the circle, hence \Sigma F=\dfrac{mv^2}{l}=mg+T \Rightarrow \dfrac{mv^2}{l} \geq mg \Rightarrow v^2 \geq gl \Rightarrow u^2=v^2+4gl \geq gl+4gl=5gl \Rightarrow u \geq \sqrt{5gl}.

    Surely v>0 for compete circular motion rather than v>gl, I'm sorry I'm confused to see where you got that from
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    (Original post by jjsnyder)
    Here are my answers from todays exam, not all of them will necessarily be correct so please correct me if you think they are wrong!
    1. a) 9.6
    b) i) 24.3
    ii) 12.7
    iii) No Air Resistance
    2.
    a)  a = (8-4x^3)i - 18e^{-3x}j
    b) i)  F = (16-8x^3)i - 36e^{-3x}j
    ii) 8.20
    c)  r = (4x^2- \frac{x^5}{5} + 3)i - (3+ 2e^{-3x})j
    3. a) By symmetry
    b)10.4 cm
    c) 16.5 degrees
    d) Explanation Q
    4. a)78.4
    b) 41.4 degrees
    c) r=2.89
    5. Lots of very long answers, will upload later if I have time. Part d) was  sqrt(5gl/2)
    6. a) Show that Q
    b)  \frac{g-(g-\lambda\mu)e^{-t\lambda} }{\lambda}
    7. a) Diagram
    b)  tanx = \frac{1-2\mu^2}{4\mu}
    8. a) 17.0
    b) I got 6.05m but many others got 7.58m so this seems like the correct answer, can anyone confirm this?
    Agree with all of these except 8 wasn't split in to a) and b) was it? Thought it was just one 8 marker? Also got 7.5m for 8 BUT forgot the cos30 when I was finding friction, but cos30 is about 0.86 so my term for friction wouldn't have been too far out, so I think 7.58 is probably right.

    Also got 2root(ag) for for 5, surely you've got 0.5mv^2 > 0.5mu^2 - 2 ag, so if v > 0 for complete revolution, u must be > root(4ag) i.e. 2root(ag)?
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    Q4a was 10.4 im pretty sure
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    (Original post by benjammy)
    Surely v>0 for compete circular motion rather than v>gl, I'm sorry I'm confused to see where you got that from
    I'm with you, the condition for continued circular motion is v>0 at the top isn't it?
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    (Original post by benjammy)
    Surely v>0 for compete circular motion rather than v>gl, I'm sorry I'm confused to see where you got that from
    Because if v<gl then the tension is negative which is impossible.
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    (Original post by IrrationalRoot)
    Guys, it's \sqrt{5gl} since we have u^2=v^2+4gl and we require T \geq 0 at the top of the circle, hence \Sigma F=\dfrac{mv^2}{l}=mg+T \Rightarrow \dfrac{mv^2}{l} \geq mg \Rightarrow v^2 \geq gl \Rightarrow u^2=v^2+4gl \geq gl+4gl=5gl \Rightarrow u \geq \sqrt{5gl}.
    Yeah, but if you go off the basis that to make revolutions, v>0 this gives min u = root (4gl). This question has come up before, and the answer was definitely 2root(gl) then.


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    (Original post by benjammy)
    Surely v>0 for compete circular motion rather than v>gl, I'm sorry I'm confused to see where you got that from
    The question you are talking about had no tension so u need the velocity to be greater than 0 but in the exam there was a string so need to the tension greater than 0
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    7.58 isn't right for the last question because there is work done against friction
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    (Original post by CoIIins)
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    this question dosen't account for tension in the wire
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    Got all the same, except 7.58 for Q8 and 2root(gl) for the other one, I may be wrong though
 
 
 
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