The Kinetic Energy K Of A Rotating Body Depends On Its Moment Of Iner

1.	The Kinetic Energy K Of A Rotating Body Depends On Its Moment Of Inertia I And Its Angular Speed ω. Considering The Relation To Be K = Kiaωb Where K Is Dimensionless Constant. Find A And B. Moment Of Inertia Of A Spehere About Its Diameter Is (2/5)mr2?
Answer» DIMENSIONS of Kinetic ENERGY K = [M1L2T-2] Dimensions of MOMENT of Inertia (I) = [ (2/5)Mr2] = [ML2T0] Dimensions of angular speed ω = [θ/t] = [M0L0T-1] Applying principle of HOMOGENEITY in dimensions in the equation K = kIaωb [M1L2T-2] = k ( [ML2T0])a([M0L0T-1])B [M1L2T-2] = k [MaL2aT-b] ⇒ a = 1 and b = 2 ⇒ K = kIω2 Dimensions of Kinetic energy K = [M1L2T-2] Dimensions of Moment of Inertia (I) = [ (2/5)Mr2] = [ML2T0] Dimensions of angular speed ω = [θ/t] = [M0L0T-1] Applying principle of homogeneity in dimensions in the equation K = kIaωb [M1L2T-2] = k ( [ML2T0])a([M0L0T-1])b [M1L2T-2] = k [MaL2aT-b] ⇒ a = 1 and b = 2 ⇒ K = kIω2

The Kinetic Energy K Of A Rotating Body Depends On Its Moment Of Inertia I And Its Angular Speed ω. Considering The Relation To Be K = Kiaωb Where K Is Dimensionless Constant. Find A And B. Moment Of Inertia Of A Spehere About Its Diameter Is (2/5)mr2?

Answer»

DIMENSIONS of Kinetic ENERGY K = [M1L2T-2]

Dimensions of MOMENT of Inertia (I) = [ (2/5)Mr2] = [ML2T0]

Dimensions of angular speed ω = [θ/t] = [M0L0T-1]

Applying principle of HOMOGENEITY in dimensions in the equation K = kIaωb

[M1L2T-2] = k ( [ML2T0])a([M0L0T-1])B

[M1L2T-2] = k [MaL2aT-b]

⇒ a = 1 and b = 2

⇒ K = kIω2

Dimensions of Kinetic energy K = [M1L2T-2]

Dimensions of Moment of Inertia (I) = [ (2/5)Mr2] = [ML2T0]

Dimensions of angular speed ω = [θ/t] = [M0L0T-1]

Applying principle of homogeneity in dimensions in the equation K = kIaωb

[M1L2T-2] = k ( [ML2T0])a([M0L0T-1])b

[M1L2T-2] = k [MaL2aT-b]

⇒ a = 1 and b = 2

⇒ K = kIω2

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