Question

A flywheel of moment of inertia I is set into rotation by the descent of a mass m. if w is the angular velocity of the flywheel and v is the linear velocity of mass when it has descended through a distance h, mgh=1/2 mv^2 + 1/2 Iw^2. show by dimentions that this equation is correct.​

Answer 1

we know,

dimension of mass , m = [M]

dimension of acceleration due to gravity, g = [LT-¹]

dimension of height, h = [L]

dimension of velocity, v = [LT-¹]

dimension of moment of inertia, I = [ML²]

dimension of angular velocity, w = [T-¹]

given equation is ..

mgh = 1/2 mv² + 1/2 Iw²

from dimensional analysis, equation is correct if and only if

dimension of mgh = dimension of {1/2mv² } = dimension of {1/2 Iw²}

so, dimension of mgh = [M][LT-²][L] = [ML²T-²]

dimension of {1/2mv²} = [M][LT-¹]² = [ML²T-²]

dimension of {1/2 Iw²} = [ML²][T-¹]² = [ML²T-²]

here it is clear that,

dimension of mgh = dimension of {1/2mv²} = dimension of {1/2 Iw²}

so, given equation is dimensionally correct.