Question

A wheel of mass 5 kg and radius 0.40 m is rolling on a road without sliding with angular velocity 10 rad s-1. The moment of inertia of the wheel about the axis of rotation is 0.65 kg m2. The percentage of kinetic energy of rotation in the total kinetic energy of the wheel is

• A. 22.4%
• B. 11.2%
• C. 88.8 %
• D. 44.8 %

Answer 1

Answer: (D)

44.8 %

Answer 2

Here, m = 5 kg, I = $0.65$ kg $\mathrm { m } ^ { 2 }$

$\omega = 10 \mathrm { rads } ^ { - 1 } , \mathrm { R } = 0.40 \mathrm {~m}$

Linear velocity,

$\mathrm { v } = \mathrm { R } \omega = 0.40 \times 10 = 4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$

Translation KE,

$= \frac { 1 } { 2 } \mathrm { mv } ^ { 2 } = \frac { 1 } { 2 } \times 16 = 40 \mathrm {~J}$Rotational KE,

$= \frac { 1 } { 2 } \mathrm { I } \omega ^ { 2 } = \frac { 1 } { 2 } \times 0.65 \times 100 = 32.5 \mathrm {~J}$

Total KE = Translational KE + Rotational KE,

$= 40 + 32.5 = 72.5 \mathrm {~J}$

Percentage of rotational KE,

$= \frac { \text { Rotational KE } } { \text { Total KE } } \times 100$

$= \frac { 32.5 } { 72.5 } \times 100 = 44.8 \%$