A circular ring of mass m and radius r is rolling on a smooth horizontal surface with speed v. Its kinetic energy is

Question

A circular ring of mass m and radius r is rolling on a smooth horizontal surface with speed v. Its kinetic energy is (A) (1/8)mv2 (B) (1/4)mv2 (C) (1/8)m2v (D) mv2

Tardigrade Admin · Accepted Answer

KE of rotation = (1/2)(mr2/2)× ω2 =(1/2)mr2× (v2/r2) =(1/2)mv2

Q. A circular ring of mass m and radius r is rolling on a smooth horizontal surface with speed v. Its kinetic energy is

$\frac{1}{8} m v^{2}$

$\frac{1}{4} m v^{2}$

$\frac{1}{8} m^{2} v$

$m v^{2}$

KE of rotation = $\frac{1}{2} \frac{m r ^{2}}{2} \times ω^{2}$
$= \frac{1}{2} m r^{2} \times \frac{v ^{2}}{r ^{2}}$
$= \frac{1}{2} m v^{2}$

Q. A circular ring of mass m and radius r is rolling on a smooth horizontal surface with speed v. Its kinetic energy is

81​mv2

41​mv2

81​m2v

mv2

KE of rotation = 21​2mr2​×ω2 =21​mr2×r2v2​ =21​mv2

$\frac{1}{8} m v^{2}$

$\frac{1}{4} m v^{2}$

$\frac{1}{8} m^{2} v$

$m v^{2}$

KE of rotation = $\frac{1}{2} \frac{m r ^{2}}{2} \times ω^{2}$
$= \frac{1}{2} m r^{2} \times \frac{v ^{2}}{r ^{2}}$
$= \frac{1}{2} m v^{2}$