A sphere of radius 1.59 cm and a spherical shell of radius 7.72 cm are rolling without slipping along the same floor. The two objects have the same mass. If they are to have the same total kinetic energy, what should the ratio of the sphere's angular speed to the spherical shell's angular speed be?

4.86

Explanation:

sphere moment of Inertia Iₑ = (2/5)mrₑ²

Let the sphere of radius 1.59 cm be x

Let the spherical shell of radius 7.72 cm be y, so that

Iₑ(x) = 2/5 * m * 1.59²

Iₑ(x) = 2/5 * m * 2.5281

Iₑ(x) = 1.011m

Iₑ(y) = 2/5 * m * 7.72²

Iₑ(y) = 2/5 * m * 59.5984

Iₑ(y) = 23.84m

Also, the angular speed of the sphere's would be ωₑ(x) and ωₑ(y)

total k.e = rotational k.e + linear k.e

for sphere = ½Iₑωₑ² + ½mωₑ²rₑ²

For sphere x

{ωₑ²[ 1.011 + 1.59²]} =

ωₑ²(1.011 + 2.5281) =

ωₑ²(3.5391)

For sphere y

{ωₑ²[ 23.84 + 7.72²]} =

ωₑ²(23.84 + 59.5984) =

ωₑ²(83.4384)

If the ratio of x/y = 1, then

ωₑ(x)²(3.5391) / ωₑ(y)²(83.4384) = 1

ωₑ(x)²(3.5391) = ωₑ(y)²(83.4384)

[ωₔ(x)/ωₑ(y)]² = [83.4384] / [3.5391] ~= 23.5762

[ωₔ(x)/ωₑ(y)] = √(23.5762)

[ωₔ(x)/ωₑ(y)] = 4.86

increasing transnational kinetic energy. because the transnational kinetic energy formula consists of only two variables, mass and velocity, increasing one of those properties is the only way to increase an object's transnational kinetic energy. increases to mass and velocity, however, do not have the same impact.

i am pretty sure it is c because when people are closer together, they cannot move as freely. picture being in a mostly full elevator, for instance. the people are right next to you, and you can barely move.