? asked in Science & MathematicsPhysics · 10 years ago

A marble of mass m and radius r rolls along the looped rough track of the figure. Ignore frictional losses....?

A marble of mass m and radius r rolls along the looped rough track of the figure. Ignore frictional losses.

a) What is the minimum value of the vertical height h that the marble must drop if it is to reach the highest point of the loop without leaving the track? Assume r << R.

b)What is the minimum value of the vertical height h that the marble must drop if it is to reach the highest point of the loop without leaving the track? Do not make assumption that r << R. Express your answer in terms of R and r.

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  • 10 years ago
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    R denotes Radius of loop, r denotes radius of sphere

    a) Conservation of Energy: Initially is at initial height h; finally is at height of loop

    PEi = PEf + KEf + Krot-f

    mgh = mg(2R) + 1/2mv^2 + 1/2(I)(w^2)

    --> I (moment of Inertia of sphere) = 2/5mr^2

    --> v=r*w (omega denotes angular velocity) therefore w=v/r

    --> at top of loop, Normal force (N) and mg pointing down with centripetal acceleration

    mg+N=mv^2/R but just making the loop means N=0 Therefore, mg=mv^2/R or v^2 (which we need)= gR

    plug everything in...

    mgh = mg(2R) + 1/2m(gR) + 1/2(2/5mr^2)(v/r)^2

    mgh = mg(2R) + 1/2m(gR) + 1/2(2/5mr^2)([gR]/r^2)

    Cancel the mg in all terms and r^2 in the last term

    h = 2R+1/2R +2/10R = 2.7R

    b) Same Idea except you must assume r contributes to potential energy and kinetic energy

    Conservation of Energy: PEi = PEf + KEf + Krot-f

    mg(h+r) = mg(2R-r) + 1/2mv^2 + 1/2(I)(w^2)

    --> I (moment of Inertia of sphere) = 2/5mr^2

    --> v=r*w (omega denotes angular velocity) therefore w=v/r

    --> at top of loop, Normal force (N) and mg pointing down with centripetal acceleration

    mg+N=mv^2/(R-r) but just making the loop means N=0 Therefore, mg=mv^2/(R-r) or v^2 (which we need)= g(R-r)

    plug everything in...

    mg(h+r) = mg(2R-r) + 1/2mg(R-r) + 1/2(2/5mr^2)(v/r)^2

    mgh = mg(2R) + 1/2m(gR) + 1/2(2/5mr^2)(g[R-r]/r^2)

    Cancel the mg in all terms and r^2 in the last term

    h+r = 2R-r+1/2R-1/2r +2/10R- 2/10r

    h= 2.7R- 2.7r

    Source(s): mastering physics UCLA go bruins!!!
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