Dynamics: planar kinetics of a rigid body (impulse and momentum)

[silentwf]

Homework Statement

http://img340.imageshack.us/img340/7379/87835960.png
$r_{i}=.3 m \text{ and } r_{o}=.9 m$
The spool has a weight of 300N and a radius of gyration $k_{O}$ = .45 m. A cord is wrapped around its inner hub and the end subjected to a horizontal force P = 50N. Determine the spool's angular velocity in 4seconds starting from rest. Assume the spool rolls without slipping.

Homework Equations

$\int_{0}^{t}{P\cdot r_{i}}{dt} = mv_{O}r_{i} + I_{O}\omega$

The Attempt at a Solution

$\left \{ \begin{matrix}\int_{0}^{4}{50 \cdot 0.3}{dt} = \frac{300}{9.81}\cdot v_{O}\cdot0.9 + \frac{300}{9.81}\cdot 0.45^{2} \cdot \omega\\v_{O}=0.9\omega\end{matrix}$
Solving the two equations i get $\omega = 1.94 \text{rad/s}$

The book provides the solution as follows
$\left \{ \begin{matrix}<br /> 50(0.9 - 0.3)4 = \frac{300}{9.81}v_{O}\cdot 0.9+\frac{300}{9.81}\cdot.45^{2}\omega \\<br /> v_{O}=\omega\cdot0.9 \end{matrix}$
I don't get why the book does .9 - .3 for its radius, i thought it was around point O?

 

[tiny-tim]

Hi silentwf!

(have an omega: ω and an integral: ∫ )

Assume the spool rolls without slipping.
…
$\int_{0}^{t}{P\cdot r_{i}}{dt} = mv_{O}r_{i} + I_{O}\omega$
…
I don't get why the book does .9 - .3 for its radius, i thought it was around point O?

[STRIKE]You can't take moments about the centre of mass (O) when it is accelerating.

But you can always take moments about the centre of rotation (A).[/STRIKE]

If you take moments about the centre of mass (O), you need to take the torque of the friction force at A into account.

To avoid that, use A instead of O, so that the friction torque is zero.

So all you need is the torque of P about A, for which the distance is 0.6

 

[tiny-tim]

oops!

I've changed my mind …

the conclusion of my last post is still correct (that you must take moments about A, not about O), but the reasoning wasn't, so I've edited it.