Mass in relation to Tangential Force

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Homework Help Overview

The problem involves a child exerting a tangential force on a disk-shaped merry-go-round, with the goal of determining its mass based on the force applied, angular speed, and time. The subject area includes concepts of rotational dynamics and torque.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to calculate the mass using a torque equation but expresses hesitation about their solution. They consider whether converting angular speed to radians per second is necessary. Other participants question the formula used and clarify the importance of angular acceleration units.

Discussion Status

Participants are actively engaging with the problem, offering clarifications and corrections regarding the formulas and units involved. There is a productive exchange of ideas, with some guidance provided on dimensional analysis and the necessity of unit conversion.

Contextual Notes

The original poster is concerned about having limited attempts to solve the problem correctly, which adds pressure to ensure their calculations are accurate. There is also a mention of a potential typographical error in their calculations.

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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[tex]\alpha[/tex]
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.
 
Last edited:
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There is mistake in your formula.

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

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

p/s: what is '0.8087' number in your calculation? I don't get it :(
 
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?
 
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.
 
Solved for it correctly thanks to the help of both of you. Thank you both.
 

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