Recent content by vsharma88

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    How to Derive the Lagrangian for a Charged Particle in a Coulomb Potential?

    It says the charge is free to move in the horizontal (x,y) plane. Does that not mean it is only allowed to move horizontally?
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    How to Derive the Lagrangian for a Charged Particle in a Coulomb Potential?

    The question says that the charge can only move horizontally, so is it necessary to have the y component?
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    How to Derive the Lagrangian for a Charged Particle in a Coulomb Potential?

    Homework Statement A charged particle of mass m and charge q is free to move in the horizontal (x, y) plane, under the influence of the Coulomb potential due to another charge Q that is fixed at the origin. Find the Lagrangian and the differential equations of motion of the mass m, in...
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    Why Does the Load Drop When One Person Lets Go of the Plank?

    ok tiny-tim i think i got it, tell me if I am right. Fnet = F1 - Fg Moment about End \sum=N - (l/2)Fg=0 0=Iw' - (l/2)(mg) 0=(ml^2/12)w' - (l/2)(mg) w'= 6g/l Moment about center of mass \sum= N*(l/2) =Iw' * (l/2) =(ml^2/12)(6g/l)*(l/2) =(1/4)mg
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    Why Does the Load Drop When One Person Lets Go of the Plank?

    hey tiny-tim i have taken the moment about the centre: \sum\tau=0=ml^2/12*w'-mg*l/2 w'=(gl/2)*(12/l^2) =(g6)/l I am now confused what to do next, you said to take moments about the centre. Is it just : \sum\tau=0=ml^2/12*w' ? Thanks
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    Why Does the Load Drop When One Person Lets Go of the Plank?

    After the person has dropped the plank, the force becomes a torque. Fnet = N - mg = ma ma = (ml^2)/12 * w' - mg [(ml^2)/2 - for a thin uniform plank] I am now stuck on what to do with w'.
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    Why Does the Load Drop When One Person Lets Go of the Plank?

    Homework Statement Two People are holding the ends of a uniform plank of lenth, l and mass, m. Show that if one person suddenly let's go, the load suppored by the other person suddenly drops from mg/2 to mg/4. Homework Equations Fnet=ma L=Iw' I=ml^2+I' The Attempt at a Solution...
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    Relative Velocity of a Helicopter

    Homework Statement Average Wind Velocity 38km/h [25 degrees N of E] Helicopter needs to achieve 91km/h [17 degrees W of N] Homework Equations r=sqrt(x^2+y^2) The Attempt at a Solution y=38sin25 + 91sin98 = 106.17 x=38cos25 - 91cos98 = 47.1 r= 116.15 answer should be 94km/h
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    Max Amplitude of Underdamped Oscillation

    Hey all, I am having trouble determing how to get the max amplitude of an underdamped oscillation. Can someone please tell me where do I start form.
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