Question

Jacob and Sophia are playing with a merry-go-round on a playground.The merry-go-round can be modeled as a flat disk with massM= 223.0kg and radiusR= 3.3 m that freely rotates horizontally without frictionabout its center axis. When the merry-go-round is already spinning at23 rotations per minute, Jacob exerts a constant forceFJ= 12.3 Ntangent to the outer edge of the merry go round in the direction of themerry-go-round’s rotation. At the same time Sophia puts her foot ata distancer= 3.1 m from the center, which exerts a constant force offrictionFS= 21.2 N. How long will it take for the merry-go-round tocome to a stop?

Answer 1

Answer:

116 s

Explanation:

We can convert the 23 rotation per minute to radian per second knowing each rotation is 2π and each minute has 60 seconds.

$\omega_0 = 2\pi*23/60 = 2.41 rad/s$

The torque generated by Jacob at the outer rim in the direction of motion

$T_J = F_JR_J = 12.3 * 3.3 = 40.59 Nm$

The torque generated by Sophia at r = 3.1 m that hampers the motion is

$T_S = F_Sr = 21.2*3.1 = 65.72 Nm$

The net torque is

$T = T_S - T_J = 65.72 - 40.59 = 25.13 Nm$

The moment of inertia of the solid disk merry-go-round is:

$I = mR^2/2$

Where m = 223 kg is the disk mass and R = 3.3 m is the radius of the disk.

$I = 223*3.3^2/2 = 1214.235 kgm^2$

So the angular deceleration is

$\alpha = T / I = 25.13 / 1214.235 = 0.0207 rad/s^2$

If the initial angular speed is 2.41, the time it'd take to decelerate to rest is

$t = \Delta \omega / \alpha = \frac{0 - 2.41}{-0.0207} = 116 s$