Question

A 357 kg merry-go-round in the shape of a horizontal disk with a radius of 2 m is set in motion by wrapping a rope about the rim of the disk and pulling on the rope. How large a torque would have to be exerted to bring the merry-go-round from rest to an angular speed of 4.3 rad/s in 2.8 s

Answer 1

τ = 1096.5 N.m

Step-by-step explanation:

given,

mass = 357 Kg

radius of the merry-go-round = 2 m

angular speed = 4.3 rad/s

time = t = 2.8 s

moment of inertia =

`I = (mr^2)/(2)`

`I = (357* 2^2)/(2)`

I = 714 kg.m²

τ = F r = I α

α =
`(\omega_f-\omega_o)/(t)`

α =
`(4.3-0)/(2.8)`

α = 1.536 rad/s²

τ = I α

τ =714 x 1.536

τ = 1096.5 N.m

Answer 2

Torque,
`\tau=1096.5\ N-m`

Step-by-step explanation:

It is given that,

Mass of the merry go round, m = 357 kg

Radius of the horizontal disk, r = 2 m

Initial speed of the merry go round,
`\omega_i=0`

Final angular speed,
`\omega_f=4.3\ rad/s`

Time, t = 2.8 s

It is set in motion by wrapping a rope about the rim of the disk and pulling on the rope. We need to find how large a torque would have to be exerted to bring the merry-go-round from rest to an angular speed of 4.3 rad/s in 2.8 s. It is given by :

`\tau=I* \alpha`

I is the moment of inertia of the disk,
`I=(mr^2)/(2)`

`\alpha` is the angular acceleration

`\tau=(mr^2)/(2)* (\omega_f-\omega_i)/(t)`

`\tau=(mr^2)/(2)* (\omega_f)/(t)`

`\tau=(357* 2^2)/(2)* (4.3)/(2.8)`

`\tau=1096.5\ N-m`

So, the torque exerted is 1096.5 N-m. Hence, this is the required solution.