Inertia block on inclined surface problem? help with solution?

[nchin]

A wheel of radius 0.20 m is mounted on a frictionless horizontal axis. A massless cord wrapped around the wheel is attaches to a 2.0 kg block that slides on a horizontal frictionless surface inclined at angle θ = 20 degrees with the horizontal. The box accelerates down the surface at 2.0 m/s^(2). What is the rotational inertia of the wheel about the axle?

Solution and picture is on page 32-33
http://mrlee.dreamstation.com/apPhysics/lect/apPhysics_lec_10.pdf

I understand all the steps except the last one where

m1(gsinθ - a1)r2 / a1 / r2 = m1(gsinθ - a1)r2^(2) / a1

why did they divide the denominator a1 by r2? and how did the r2 in the numerator become r2^(2)??


thanks!

 

[grzz]

Note that linear acceleration of the string wound round the pulley is the same as a$_{1}$ since the string is assumed that does not extend.

Now linear velocity = radius x angular velocity.

Similarly linear acceleration = radius x angular acceleration.

i.e. a$_{2}$ = r$_{2}$α$_{2}$

but a$_{2}$ = a$_{1}$

therefore a$_{1}$ = r$_{2}$α$_{2}$

i.e. α$_{2}$ = $\frac{a_{1}}{r_{2}}$

 

[Doc Al]

and how did the r2 in the numerator become r2^(2)??

1/(1/x) = x

x/(1/x) = x^2

 

[nchin]

Note that linear acceleration of the string wound round the pulley is the same as a$_{1}$ since the string is assumed that does not extend.

Now linear velocity = radius x angular velocity.

Similarly linear acceleration = radius x angular acceleration.

i.e. a$_{2}$ = r$_{2}$α$_{2}$

but a$_{2}$ = a$_{1}$

therefore a$_{1}$ = r$_{2}$α$_{2}$

i.e. α$_{2}$ = $\frac{a_{1}}{r_{2}}$

Then how did r2 become r2 ^ (2)?

 

[grzz]

As Doc Al told you.

The denomenator of a denomenator is a numerator.

 

[nchin]

Whoops I thought the alpha was a regular a.

Thanks!