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

Answer:

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

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 :

$\tau=I\times \alpha$

I is the moment of inertia of the disk, $I=\dfrac{mr^2}{2}$

$\alpha$ is the angular acceleration

$\tau=\dfrac{mr^2}{2}\times \dfrac{\omega_f-\omega_i}{t}$

$\tau=\dfrac{mr^2}{2}\times \dfrac{\omega_f}{t}$

$\tau=\dfrac{357\times 2^2}{2}\times \dfrac{4.3}{2.8}$

$\tau=1096.5\ N-m$

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

Answer 2

Answer:

τ = 1096.5 N.m

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 =

$I = \dfrac{mr^2}{2}$

$I = \dfrac{357\times 2^2}{2}$

     I = 714 kg.m²

τ = F r  = I α

α = $\dfrac{\omega_f-\omega_o}{t}$

α = $\dfrac{4.3-0}{2.8}$

α = 1.536 rad/s²

τ = I α

τ =714 x 1.536

τ = 1096.5 N.m