# Finding the equation of motion regarding a block pulling a uniform disk [closed]

My problem is finding the equation of motion for the block. Using Newton's Second Law, I did: $$4g-T-R=4\frac{dw}{dt}$$ But the mark scheme says: $$4g-T-R=0.2*4\frac{dw}{dt}$$ And I don't understand why the radius has anything to do with this. Can somebody please explain?

• Consider the Angular Momentum. Commented Jun 4, 2014 at 16:34
• But that doesn't have to do with the second law, right? Commented Jun 4, 2014 at 16:49
• ccrma.stanford.edu/~jos/pasp/Newton_s_Second_Law_Rotations.html Commented Jun 4, 2014 at 16:52

$$$\begin{cases} Mg - T - R = Ma \\ Tr = I\alpha \end{cases}$$$

First equation from Secondo Law of Dynamics, second equation is $\sum M_i = I \frac{d^2\theta}{dt^2}$ (Euler's second Law). Since $I = Mr^2/2$, and $\alpha r = a$(then $\alpha = \frac{d\omega}{dt} \Leftarrow a = r\frac{d\omega}{dt}$, that is your error!) because the string is running with same velocity of the side of your disk, you obtain

$$$\begin{cases} T = M(g - a) - R \\ Mgr - Mar - Rr = Mr/2(r\alpha) = Mra/2 \end{cases}$$$

$Mg - Ma - R = Ma/2$

$a = 2\frac{g - R/M}{3}$

$\alpha = 2\frac{g - R/M}{3r}$

from $\alpha= \frac{\omega^2}{2\theta}$ you obtain the needed value of $\alpha$, substiting in the previous equation you get $T$ and $R$.