Question

A 150-kg merry-go-round in the shape of a uni- form, 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 must be exerted on the rope to bring the merry-go-round from rest to an angular speed of 0.500 rev/s in 2.00 s?

Answer 1

The constant force that must be exerted on the rope to bring the merry-go-round from rest to the angular speed is 2,218.5 N.

How to calculate the constant force exerted on the rope?

The constant force that must be exerted on the rope to bring the merry-go-round from rest to an angular speed of 0.500 rev/s in 2.00 s is calculated as follows;

F = ma

where;

• m is the mass
• a is the centripetal acceleration

a = ω²r

where;

• ω is the angular speed
• r is the radius

ω = 0.5 rev/s x 2π rad/rev

ω = 3.14 rad/s

a = (3.14²) x 1.5 m

a = 14.79 m/s²

The force that must be applied is;

F = ma

F = 150 kg x 14.79 m/s²

F = 2,218.5 N

Answer 2

The force required to pull the merry-go-round is 241.9 N

Step-by-step explanation:

ω = ωo + a×t

where ω is the angular speed, t is time, ωo is the initial angular speed

0.7 rev / s ×(2×π radians/rev) = 4.3 radians/s

4.3 = a×2

a = 2.15 rad/s^2

then:

F×r = I×a

Where F is the force on the merry-go-round, r is the radius, I is the moment of inertia of the merry-go-round, and a is the angular acceleration

for a solid, horizontal disk, the moment of inertia is: 1/2×m×r^2

that is:

F×r = 1/2×m×r^2×a

F = 1/2×m×r×a

= 1/2×(150)×(1.50)×(2.15)

= 214.9 N