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Paymemoney
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Homework Statement
A playground merry-go-round of radius R=2.00m has a moment of intertia [tex]I=250kg.m^2[/tex] and is rotating at 10 rev/min (rpm) about a frictionless vertical axle. Facing the axle, a 25.0kg child hops onto the merry-go-round from the ground and manages to sit down on its edge. What is the new angular speed of the merry-go-round?
Homework Equations
[tex]\tau=rF[/tex]
[tex]\tau=I\alpha[/tex]
[tex]\omega final = \omega initial + \alpha t[/tex]
The Attempt at a Solution
I have done this, however my answer is incorrect.
[tex]\tau=rF[/tex]
[tex]\tau=2*25a[/tex]
[tex]\tau=50a[/tex]
[tex]\tau=Im^2[/tex]
[tex]50a=250\alpha[/tex]
[tex]v=\frac{x}{t}[/tex]
[tex]1.047=\frac{2}{t}[/tex]
[tex]t=2.094[/tex]
so...
[tex]50*{\omega}{t}=250\alpha[/tex]
[tex]50*\frac{1.047}{2.094}=250\alpha[/tex]
[tex]\alpha=\frac{25}{250}[/tex]
[tex]\alpha=0.1[/tex]
now to find the final angular speed
[tex]\omega final = \omega initial + \alpha t[/tex]
[tex]\omega final = 1.047 + 0.1*2.094[/tex]
[tex]\omega final = 1.2564rad/s[/tex]
Also a quick question can someone explain to me what is the difference between the
[tex]\tau=rF[/tex] & [tex]\tau=I\alpha[/tex] equations??
I'm not quite sure if this is correct; Is the [tex]\tau=rF[/tex] normally used for small rotating mass, and [tex]\tau=I\alpha[/tex] is used for a large rotating body.
P.S
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