Recent content by saket

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    Conceptual Question on Kinetic Friction

    yeah right.. and what we do normally use is, Newtonian assumption (for dry friction): µk & fk doesnt depend upon surface area, or speed.. (µk doesnt depend upon even mass) ..but then it is just an assumption.. for we to be able to work easily, and possibly it matches agreeably (although not...
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    Conceptual Question on Kinetic Friction

    If I am not mistaken following are the findings which some standard text books have quoted: µk, for rigid bodies, depends on the two materials in contact and is independent of mass, m. Therefore, in a given problem fk is proportional to normal reaction (as fk = µkN). But also, some have quoted...
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    Static Equilibrium and Balance

    Something amiss in the question.. or, there shouldn't be (2) part.
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    Conservation of EnErGy Problem help please

    I think your statements have flaws. Have you directly quoted the question or have you tried to put it up in your own words? And, when you say, "i got the wrong answer" .. you should be knowing the correct answer? Anyways, as far as classical Kinematics go: Kinetic Energy = m*(v^2)/2...
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    Rotating Pail of Water

    Method I: Ideal Fluid, by definition, can not tolerate any shear force. Use this fact. Assume the centre of the water surface to be origin. Let P(x,y) be a point on the water surface. Consider forces on it. Impose the condition that tangential force will be zero. I hope you can proceed now.
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    Conservation of EnErGy Problem help please

    well.. 11.2 * 1000 = 11200 and not 112000!! what is the combined mass?? 105kg? or, 2,3105kg, or what?? what is the answer?
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    What was Dan’s final velocity at this time?

    (b) seems fine.. (assuming constant acceleration) why dont you try for (a) in a similar fashion!
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    Torque and Rotational Kinematics

    Okay, although not provided in the question, I am assuming that you are supposed to find coefficient of kintic friction. (Your attempt at the solution tells me that.) You erred in this statement: Note that, torque produced by F is not the one which is producing deceleration in the disc (It...
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    Work done by Gravity on a liquid

    Consider a stick of length L. Where will its centre of mass be? Mid-point, right? This is at a distance of L/2 from either end. Similarly in the given problem, we have water column of height, say h. So, centre of mass will be at a height of h/2 .. isnt it? {You have studied centre of mass...
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    Work done by Gravity on a liquid

    Hmm.. gone to sleep? Hope, you will solve it in your dreams! Anyways, if in case you are not able to, to make you start your day happily, here is the soltion: (Both methods are almost same.) Method I: Let bottom of the cylinders be our reference line for height. Let h1 = 1.56 m, h2 = 0.854 m...
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    Work done by Gravity on a liquid

    No, you wont need integral.
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    Trouble Determining Moment of Inertia Formula

    Wait, I am a bit confused about the figure. First thing, your wedge is a right-triangular prism, of some thickness, t. The triangular face has sides L and h. (None of these are hypotenuse of the triangle.) Now let us consider this triangular face on the plane of the paper (such that...
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    Trouble Determining Moment of Inertia Formula

    If that is the case, then don't you think, your following formula doesnt hold here: (Why?) Simply because, it is a thin rod (of thickness dx) and length -- that can be obtained using similar triangles (or trigonometry). So, you should use moment of inertia for a thin rod (about its centre)...
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    Trouble Determining Moment of Inertia Formula

    yep, is it okay? (Attachment is still in pending for approval. :( )
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    Work done by Gravity on a liquid

    No no, you dont need to go for Bernoulli. Note that as liquid is enetring/leaving the cylinder, mass of the liquid in it doesnt remain constant. So, the formula you have used for change in potential energy is wrong. Instead, use \DeltaPE = (mgh)final - (mgh)initial, for each of the cylinders.
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