1. Dec 22, 2013

### transparent

Tell me where I've gone wrong (here r(1,2) means position of point 1 with respect to point 2 etc):

Ï‰ x r(1,2)=v(1,2)
Differentiating both sides with respect to time:
Ï‰ x v(1,2) + $\alpha$ x r(1,2) = a(1,2)
=>Ï‰ x (Ï‰ x r(1,2)) + $\alpha$ x r(1,2) = a(1,2)

Now let us imagine a uniform rod in free space of length l and mass m, lying with the origin at its centre. A force F(+$\hat{j}$) is applied at one end (+l/2$\hat{i}$). We observe the motion about centre of mass first at time 0:
Icm=ml2/12
$\tau$=+l/2$\hat{i}$ x F$\hat{j}$=F*l/2$\hat{k}$.
$\alpha$=6F/(m*l) $\hat{k}$
Now since Ï‰=0,
a(cm, positive end)=6F/(m*l) $\hat{k}$ x l/2$\hat{i}$= 3F/m $\hat{j}$

Now we observe the motion about the positive end:
$\tau$=0 (since the force is applied at that very point)
=> $\alpha$=0
Ï‰=0
Hence a(positive end, cm)=0
=>a(cm, positive end)=0

What have I done wrong? :(
Edit: By positive end I mean the end which is on the positive side of the x axis.

Last edited: Dec 22, 2013
2. Dec 22, 2013

### Staff: Mentor

Things aren't so simple when you take torques with respect to an accelerating point other than the center of mass. You cannot set net torque equal to IÎ± in such a case.

3. Dec 22, 2013

### transparent

Why? And isn't the centre of mass accelerated?

Edit: Didn't read "other than the centre of mass" part. But still, why is it so?