What is the new angular speed of the merry-go-round after a child hops on?

[Paymemoney]

Homework Statement

A playground merry-go-round of radius R=2.00m has a moment of intertia I=250kg.m^2 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

\tau=rF

\tau=I\alpha

\omega final = \omega initial + \alpha t

The Attempt at a Solution

I have done this, however my answer is incorrect.

\tau=rF

\tau=2*25a

\tau=50a

\tau=Im^2

50a=250\alpha

v=\frac{x}{t}

1.047=\frac{2}{t}

t=2.094

so...
50*{\omega}{t}=250\alpha

50*\frac{1.047}{2.094}=250\alpha

\alpha=\frac{25}{250}

\alpha=0.1

now to find the final angular speed
\omega final = \omega initial + \alpha t

\omega final = 1.047 + 0.1*2.094

\omega final = 1.2564rad/s

Also a quick question can someone explain to me what is the difference between the
\tau=rF & \tau=I\alpha equations??
I'm not quite sure if this is correct; Is the \tau=rF normally used for small rotating mass, and \tau=I\alpha is used for a large rotating body.

P.S

 

[tiny-tim]

Hi Paymemoney!

(have a tau: τ and an omega: ω and an alpha: α and try using the X² tag just above the Reply box )

A playground merry-go-round of radius R=2.00m has a moment of intertia I=250kg.m^2 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

\tau=rF …

No … "facing the axle" mans that there is no external torque …

start again, using only conservation of angular momentum. ​

Also a quick question can someone explain to me what is the difference between the
\tau=rF & \tau=I\alpha equations??
I'm not quite sure if this is correct; Is the \tau=rF normally used for small rotating mass, and \tau=I\alpha is used for a large rotating body.

τ = r x F tells you how much the torque is

τ = Iα tells you what the torque does

(and so r x F = Iα tells you what the force does)

 

[Paymemoney]

When you are talking about what the torque does, do you mean if it slowing down or accelerating?

 

[tiny-tim]

more than that …

i mean that, just as (linear) force F tells you the (linear) acceleration a (of a body with mass m),

torque τ tells you the angular acceleration α (of a body with moment of inertia I)

… they both tell you exactly what the force or torque does to the body ​