Torque's lever arm to attain specific accel of center of mass

Question

I have trouble with the following exersice.

A disc of mass $M$ and radius $R$ rolls without slipping.

$F=10N$

$M=5kg$

$I=MR^2/2$

$a_{cm}=2,5 m/s^2$

So, I tried the following:

$$\sum F_x = F-F_\mu$$ $$\tau=F·sin(\theta)R+F_\mu·R$$ Get an expression for $F_\mu$ $$a_{cm}=\frac{\sum F_x}{M}=\frac{F-F_\mu}{M} \Rightarrow F_\mu=-a_{cm}M+F$$ Then substitute in the torque equation: $$\tau=F·sin(\theta)R+(-a_{cm}M+F)·R $$ $$\tau= R(F·sin(\theta)-a_{cm}M+F)$$ Now, get angular acceleration from torque: $$\alpha = \frac{\tau}{I} = \frac{2R}{MR^2}(F·sin(\theta)-a_{cm}M+F)$$ $$\alpha = \frac{2}{MR}(F·sin(\theta)-a_{cm}M+F)$$ Now get tangencial acceleration: $$a_t=\alpha R$$ $$a_t= \frac{2}{M}(F·sin(\theta)-a_{cm}M+F)$$ Equate to $a_{cm}$ $$a_{cm}=a_t/2$$ $$a_{cm}=\frac{F·sin(\theta)-a_{cm}M+F}{M}$$ Now solve for $sin(\theta)=h/R$ $$2a_{cm}=\frac{F·sin(\theta)+F}{M}$$ $$\frac{2a_{cm}M}{F}-1 = sin(\theta) = h/R$$ $$h=(\frac{2a_{cm}M}{F}-1)R = (\frac{(2)(2,5)(5)}{10}-1)R$$ $$h=1,5R$$

And that is obviously wrong.

Where is the mistake? Thanks!

Answer 1

A much simpler approach is to consider the moment of inertia about the "instantaneous point of rotation" - namely the edge of the disk.

From the parallel axes theorem, the moment of inertia about this point (distance $R$ from the center of mass) is

$$I_e = I_c + MR^2 = \frac32 M R^2$$

Now you want a linear acceleration of $a$, which implies an angular acceleration of $\dot\omega = \frac{a}{R}$.

Finally, the torque applied is $\Gamma = F\cdot(R+h)$.

The equation of motion is

$$\Gamma = I_e \dot\omega$$

Substitute the expressions for each of the three terms, and solve for $h$. I think you will find it a faster approach. You are making life harder for yourself by trying to carry $\theta$ forward in the equations, when the torque can be found directly (this is possible because there is no slip...).