Question

A 160-kg merry-go-round in the shape of a uniform, solid, horizontal disk of radius 1.50 m is set in motion by wrapping a rope about the rim of the disk and pulling on the rope. What constant force would have to be exerted on the rope to bring the merry-go-round from rest to an angular speed of 0.600 rev/s in 2.00 s? (State the magnitude of the force.)

Answer 1

F = 453.6 N

Step-by-step explanation:

We apply Newton's second law to the merry-go-round :

∑Ft= m*at Formula (1

Where:

∑Ft :algebraic sum of tangential forces (N)

m: merry-go-round mass (Kg)

at: tangential acceleration (m/s²)

Kinematics equations of the merry-go-round

ω= ω₀ + α*t Formula (2)

at = α*R Formula (3)

Where:

α : angular acceleration (rad/s²)

t : time interval (s)

ω₀ : initial angular speed ( rad/s)

ωf : final angular speed ( rad/s)

R : radius of the circular path (m)

Data

m=160-kg

R= 1.50 m

ω₀=0

ωf = 0.600 rev/s

1 rev = 2π rad

ωf = 0.6*2π rad/s = 3.77 rad/s

t = 2 s

Calculation of angular acceleration of the merry-go-round

We replace data in the formula (2)

ω= ω₀ + α*t

3.77= 0 + α*(2 )

α= 3.77 / 2

α= 1.89 rad/s²

Calculation of tangential acceleration

We replace data in the formula (3)

at= α*R

at= (1.89)*(1.5)

at= 2.835 m/s²

Constant force (F) that would have to be exerted on the rope to bring for the merry-go-round from rest to an angular speed of 0.600 rev/s in 2.00 s :

We replace data in the formula (1)

F = m*at

F = (160 kg) * (2.835 m/s²)

F = 453.6 N