It is well known that for a hollow, cylindrical shell rolling without slipping on a horizontal surface, ?

half of the total kinetic energy is translational and half is rotational. What fraction of the total kinetic energy is rotational for the following objects rolling without slipping on a horizontal surface?

a)A uniform solid cylinder

b)A uniform sphere.

c)A thin-walled hollow sphere.

d)A hollow, cylinder with outer radius R and inner radius R/2.

3 Answers

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  • NCS
    Lv 7
    3 weeks ago

    Welcome to Yahoo!Answers.

    total KE = trans KE + rotation KE

    total KE = ½mv² + ½Iω²

    "without slipping" means that

    ω = v/R, so

    KE = ½mv² + ½I(v/R)² = (v²/2)(m + I/R²)

    If we define I = kmR², then

    I/R² = km

    and so

    KE = (v²/2)*(m + km) = (mv²/2)*(1 + k)

    This means that the fraction of the total KE that is rotational is

    → k / (1 + k)

    for a hollow cylinder, k = 1 and so we get the fraction

    1 / (1+1) = ½

    as stated.

    A) uniform solid cylinder, k = ½, so

    ½ / (1 + ½) = ½ / (3/2) = 1/3

    B) uniform (solid) sphere, k = 2/5, so

    (2/5) / (1 + 2/5) = (2/5) / (7/5) = 2/7

    C) thin hollow sphere, k = 2/3, so

    (2/3) / (1 + 2/3) = (2/3) / (5/3) = 2/5

    D) This one is trickier. People often err by finding the difference between the squares of the radii, as the other poster has done -- instead, you have to use the sum; so

    I = ½m(R² + (R/2)²) = ½mR²(1 + ¼) = (5/8)mR²

    so k = 5/8

    and

    (5/8) / (1 + 5/8) = (5/8) / (13/8) = 5/13

    If you find this helpful, please award Best Answer. You get points too!

  • 3 weeks ago

    Mistake: not "half of" but "part of" the total energy

    ....etc.

    a) For a uniform solid cylinder:

    let v=the linear speed of the object(m/s)

    .....r=the radius of the cross-section (m)

    ....T=the time needed for 1 revolution (s)

    m=the mass of the solid (kg)

    I=the moment of inertia about the axis of

    the cylinder=(mr^2)/2

    ....w=the angular speed(rad./s)

    The total kinetic energy=

    (mv^2+Iw^2)/2=

    [m(2(pi)r/T)^2+Iw^2]/2=

    [m(2(pi)r/(2pi/w))^2+Iw^2]/2=

    [m(wr)^2+Iw^2]/2

    The rotational kinetic energy=

    (w^2)[mr^2+I]/2

    =>

    The rotational k.energy/ The total k. energy=

    0.5I(w^2)/[0.5(w^2)(mr^2+0.5mr^2)]=

    0.5mr^2/[3m(r^2)/2]=1/3.

    b), c), d) can be done similarly with

    different I, it is left as your exercise.

  • Anonymous
    3 weeks ago

    a)A uniform solid cylinder

    KE = 1/2 tr.+1/4 rot = 3/4 tot

    rot/tot = 1/4 / 3/4 = 1/3 = 33%

    b)A uniform sphere.

    KE = 1/2 tr.+1/5 rot = 7/10 tot

    rot/tot = 2/10 / 7/10 = 2/7 = 28.6%

    c)A thin-walled hollow sphere.

    KE = 1/2 tr.+1/3 rot = 5/6 tot

    rot/tot = 1/3 / 5/6 = 2/5 = 40.0%

    d)A hollow, cylinder with outer radius R and inner radius R/2.

    KE = 1/2 tr.+1/2*(4-1)/4 = 3/8 rot = 7/8 tot

    rot/tot = 3/8 / 7/8 = 3/7 = 42.9%

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