# Calculate Period?

A m=43.5 kg boy is standing at the centre of a merry-go-round (a solid disk with mass M=51.5 kg and radius of R=2.99 m). The merry-go-round is turning with a period of T=6.32 s. The boy walks along a radial line to the edge of the merry-go-round. What is the period of the merry-go-round after the boy walks to the edge?

How do I solve this problem?

### 2 Answers

- billrussell42Lv 71 year agoBest Answer
Angular momentum is conserved. Calculate the angular momentum for the initial condition, L = Iω

I, moment of inertia, is that for the disk plus that of the boy, considered a point object. Initially zero as he is in the center.

now calculate the new I with the new location for the boy, as the sum of that for the diak and the boy, considered a point source, see below.

equate two angular momentums and solve for ω₂

I₁ω₁ = I₂ω₂

(you did ask for how, not for the solution)

ω = V/r

ω is angular velocity in radians/sec

1 radian/sec = 9.55 rev/min

r is radius of circle in meters

V is the tangental velocity in m/s

Angular momentum in kg•m²/s

L = Iω

I is moment of inertia in kg•m²

ω is angular velocity in radians/sec

I is moment of inertia in kg•m²

I = cMR²

M is mass (kg), R is radius (meters)

c = 1 for a ring or hollow cylinder

c = 2/5 solid sphere around a diameter

c = 7/5 solid sphere around a tangent

c = ⅔ hollow sphere around a diameter

c = ½ solid cylinder or disk around its center

c = 1/12 rod around its center, R = length

c = ⅓ for a rod around its end, R = length

c = 1 for a point mass M at a distance R from

the axis of rotation

- Some BodyLv 71 year ago
Angular momentum is conserved:

I₁ ω₁ = I₂ ω₂

(1/2 M R²) (2π rad / T₁) = (1/2 M R² + m R²) (2π rad / T₂)

(1/2 M R²) / T₁ = (1/2 M R² + m R²) / T₂

(1/2 M) / T₁ = (1/2 M + m) / T₂

M / T₁ = (M + 2m) / T₂

T₂ = T₁ (M + 2m) / M

Given T₁ = 6.32 s, M = 51.5 kg, and m = 43.5 kg:

T₂ = (6.32 s) (51.5 kg + 2 * 43.5 kg) / 51.5 kg

T₂ = 17.0 s