A thin rod is pivoted around its end,and an identical rod around its center.If they are to have the same rotational kinetic energy,then $\omega_{end}$/$\omega_{center}$ =
A) 2
B) 1
C) 1/2
D) 1/4
E) 4

Question

Quizplus · Accepted Answer

The rotational kinetic energy of a rigid body is given by:

Krot = (1/2) I ω^2

where I is the moment of inertia and ω is the angular velocity.

For a thin rod pivoted around its end, the moment of inertia is:

Iend = (1/3) ml^2

where m is the mass of the rod and l is its length.

For a identical rod pivoted around its center, the moment of inertia is:

Icenter = (1/12) ml^2

The rods have the same mass and length, so we can write:

Icenter = (1/4) Iend

To have the same kinetic energy, their angular velocities must satisfy:

(1/2) Iend ωend^2 = (1/2) Icenter ωcenter^2

ωcenter^2 = 2 ωend^2

ωcenter/ωend = √2

The ratio of the tangential forces Fend and Fcenter required to produce these angular velocities is:

Fend/Fcenter = (Icenter/Isphere) (ωend/ωcenter)

where Isphere is the moment of inertia of a sphere with the same mass and radius as the rod.

Since the rods have identical mass and length, we can set:

Isphere = (2/5) m(l/2)^2 = (1/5) ml^2

Fend/Fcenter = (Icenter/Isphere) (ωend/ωcenter)

Fend/Fcenter = (1/4) (1/√2)

Fend/Fcenter = 1/2√2

Fend/Fcenter = √2/4

Fend/Fcenter = 0.353

Fend/Fcenter ≈ 1/2

Therefore, the correct answer is C) 1/2.

A Thin Rod Is Pivoted Around Its End,and an Identical ωend\omega_{end}ωend​

A Thin Rod Is Pivoted Around Its End,and an Identical $\omega_{end}$