Question

The kinetic energy K of a rotating body depends on its
moment of inertia I and its angular speed . Assuming
the relation to be K=kId where k is a dimensionless
constant, find a and b. Moment of inertia of a sphere
about its diameter is 2/5Mr2.​

Answer 1

Given K= kIᵃωᵇ -------→ (i)

Given K= kIᵃωᵇ -------→ (i)where k is the kinetic energy of any rotating body also a dimensionless constant.

Given K= kIᵃωᵇ -------→ (i)where k is the kinetic energy of any rotating body also a dimensionless constant.So,

Given K= kIᵃωᵇ -------→ (i)where k is the kinetic energy of any rotating body also a dimensionless constant.So,Dimension of Kinetic energy, K= [ML²T⁻²]

Given K= kIᵃωᵇ -------→ (i)where k is the kinetic energy of any rotating body also a dimensionless constant.So,Dimension of Kinetic energy, K= [ML²T⁻²]Dimension of Moment of Inertia , Iᵃ = [ML²]ᵃ

Given K= kIᵃωᵇ -------→ (i)where k is the kinetic energy of any rotating body also a dimensionless constant.So,Dimension of Kinetic energy, K= [ML²T⁻²]Dimension of Moment of Inertia , Iᵃ = [ML²]ᵃDimension of angular speed , ωᵇ = [T⁻¹]ᵇ

Given K= kIᵃωᵇ -------→ (i)where k is the kinetic energy of any rotating body also a dimensionless constant.So,Dimension of Kinetic energy, K= [ML²T⁻²]Dimension of Moment of Inertia , Iᵃ = [ML²]ᵃDimension of angular speed , ωᵇ = [T⁻¹]ᵇas we studied the principle of homogeneity of dimension,

Given K= kIᵃωᵇ -------→ (i)where k is the kinetic energy of any rotating body also a dimensionless constant.So,Dimension of Kinetic energy, K= [ML²T⁻²]Dimension of Moment of Inertia , Iᵃ = [ML²]ᵃDimension of angular speed , ωᵇ = [T⁻¹]ᵇas we studied the principle of homogeneity of dimension,Now put dimension in eq (i)

Given K= kIᵃωᵇ -------→ (i)where k is the kinetic energy of any rotating body also a dimensionless constant.So,Dimension of Kinetic energy, K= [ML²T⁻²]Dimension of Moment of Inertia , Iᵃ = [ML²]ᵃDimension of angular speed , ωᵇ = [T⁻¹]ᵇas we studied the principle of homogeneity of dimension,Now put dimension in eq (i)[ML²T⁻²] = [ML²]ᵃ × [T⁻¹]ᵇ

Given K= kIᵃωᵇ -------→ (i)where k is the kinetic energy of any rotating body also a dimensionless constant.So,Dimension of Kinetic energy, K= [ML²T⁻²]Dimension of Moment of Inertia , Iᵃ = [ML²]ᵃDimension of angular speed , ωᵇ = [T⁻¹]ᵇas we studied the principle of homogeneity of dimension,Now put dimension in eq (i)[ML²T⁻²] = [ML²]ᵃ × [T⁻¹]ᵇEquating the Exponents of the similar Quantities.

Given K= kIᵃωᵇ -------→ (i)where k is the kinetic energy of any rotating body also a dimensionless constant.So,Dimension of Kinetic energy, K= [ML²T⁻²]Dimension of Moment of Inertia , Iᵃ = [ML²]ᵃDimension of angular speed , ωᵇ = [T⁻¹]ᵇas we studied the principle of homogeneity of dimension,Now put dimension in eq (i)[ML²T⁻²] = [ML²]ᵃ × [T⁻¹]ᵇEquating the Exponents of the similar Quantities.2=2a ; -b=-2    ∵ (L²=L²ᵃ & T⁻²=T⁻ᵇ)

Given K= kIᵃωᵇ -------→ (i)where k is the kinetic energy of any rotating body also a dimensionless constant.So,Dimension of Kinetic energy, K= [ML²T⁻²]Dimension of Moment of Inertia , Iᵃ = [ML²]ᵃDimension of angular speed , ωᵇ = [T⁻¹]ᵇas we studied the principle of homogeneity of dimension,Now put dimension in eq (i)[ML²T⁻²] = [ML²]ᵃ × [T⁻¹]ᵇEquating the Exponents of the similar Quantities.2=2a ; -b=-2    ∵ (L²=L²ᵃ & T⁻²=T⁻ᵇ)Hence, a=1 ; b= 2

Given K= kIᵃωᵇ -------→ (i)where k is the kinetic energy of any rotating body also a dimensionless constant.So,Dimension of Kinetic energy, K= [ML²T⁻²]Dimension of Moment of Inertia , Iᵃ = [ML²]ᵃDimension of angular speed , ωᵇ = [T⁻¹]ᵇas we studied the principle of homogeneity of dimension,Now put dimension in eq (i)[ML²T⁻²] = [ML²]ᵃ × [T⁻¹]ᵇEquating the Exponents of the similar Quantities.2=2a ; -b=-2    ∵ (L²=L²ᵃ & T⁻²=T⁻ᵇ)Hence, a=1 ; b= 2Hope it Helps.