Mass in relation to Tangential Force

[to819]

Homework Statement

A child exerts a tangential 42.4 N force on the rim of a disk-shaped merry-go-round with a radius of 2.20 m. If the merry-go-round starts at rest and acquires an angular speed of 0.0870 rev/s in 4.50 s, what is its mass?

Homework Equations

t=I\alpha
I=.5mr²

The Attempt at a Solution

42.4=.5*2.2²*.8087*m
m=21.665

Would the above be correct? I'm hesitant about entering it as I only have one more attempt to solve this problem. If not would converting the revolutions per second into radians per second be the right thing to do? Anyone see anything I'm missing? Any help is appreciated. Thank you.

 

[ApexOfDE]

There is mistake in your formula.

Firstly, torque formula is \tau = F.R.sin(\theta) (\theta is angle between F and R)

Besides, torque can be found by using this formula: \tau = I. \alpha (\alpha is angular acceleration)

p/s: what is '0.8087' number in your calculation? I don't get it :(

 

[to819]

Ah, thank you for that. That number was me mistyping 0.0870. Should I convert the angular speed to radians per second before I solve for angular acceleration or leave it as is?

 

[ideasrule]

Try doing dimensional analysis. If you convert to radians, you'll get an angular acceleration in radians/s^2. If you leave it as is, you'll get an acceleration in revs/s^2. Both are correct, but the formula torque=I*alpha requires that alpha be in radians per second squared.

 

[to819]

Solved for it correctly thanks to the help of both of you. Thank you both.