Question

How much work is required to uniformly accelerate a merry-go-round of mass 1490 kg and a radius of 7.80 m from rest to a rotational rate of 1 revolution per 8.33 s? Model the merry-go-round as a solid cylinder (I = ½ MR2). How much power is required to accelerate the merry-go-round to that rate in 8.33 s?

Answer 1

Step-by-step explanation:

Given:

Time, t = 8.33 s

Mass, M = 1490 kg

Angular velocity, w = 1 rev in 8.33 s

= 0.12 rev/s

Converting rev/s to rad/s,

1 rev = 2pi rad

0.12 rev/s × 2pi rad/1 rev

= 0.754 rad/s

Radius, r = 7.8 m

Inertia, I = 1/2 × M × r^2

= 1/2 × 1490 × 7.8^2

= 45325 kgm^2

Kinetic energy, Uk = 1/2 × I × w^2

= 1/2 × 45325 × 0.754^2

= 12884.5 J

= 12.8 kJ

Power = energy/time

Kinetic energy, Uk = energy

Power = 12884.5/8.33

= 1546.725 W

= 1.55 kW.

Answer 2

W = 12,884.221 J

Step-by-step explanation:

I = ½ MR² (given)

I = 0.5 × 1490 kg × (7.8 m)²

I = 45,325.8 Kg m²

and ω = 2π / T = 2 × 3.14 / 8.33 s

ω = 0.754 radians/s

now W = Δ K.E. = 1/2 Iω² = 0.5 × 45,325.8 Kg m² × (0.754 radians/s)²

W = 12,884.221 J