Two ropes pulling on a cylinder that is translating and rotating on a plane

[lorenz0]

Homework Statement

A homogeneous cylinder with radius $R$ and mass $M$ is resting on a plane as shown in the figure. A rope (rope $1$), which is inextensible and has negligible mass, is attached to the center of mass of the cylinder. The other end of this rope (through an ideal pulley) holds a body with a mass of $m = 50.0 kg$. A second rope (rope 2), also inextensible and of negligible mass, is wrapped around a groove of radius $r = R/2$ and has one of its ends fixed to a nearby wall (see figure).

Given that rope $2$ cannot slip relative to the groove of the cylinder and that the coefficients of static and kinetic friction between the cylinder and the supporting plane are $\mu_s = 0.300$ and $\mu_k = 0.200$, determine:

a) The values of $M$ for which the system can remain in static equilibrium.

Finally, in the case where $M = 20.0$ kg, determine:
b) The acceleration $a$ with which the hanging mass descends.

(For the calculation of the moment of inertia of the cylinder, do not take the groove into account)

Relevant Equations

$\tau=\vec{r}\times\vec{F}=I \alpha$, $\vec{F}=m \vec{a}$

I have solved the problem, but I am not sure if my solution is correct, so I would appreciate if someone would double check it. Thanks

(a) Using an inertial reference frame in which the $x$ axis points to the left and the $y$ axis points downwards and computing the torques with respect to the point of contact between the cylinder and the floor, I get:
$$
\begin{cases}
T_1-T_2-f_s=0, \\
N=Mg,\\
RT_1-(R+r)T_2=0,\\
T_1=mg
\end{cases}$$, which implies that $f_s=\frac{r}{r+R}mg \leq \mu_s Mg \Rightarrow M \geq \frac{r}{\mu_s (R+r)} m=\frac{m}{3 \mu_s}\approx 55.6 kg$

(b) Using the same frame of reference as in (a), but now computing the torques with respect to the center of mass I get:

$$
\begin{cases}
T_1-T_2-f_d=M a, \\
N=Mg,\\
Rf_d-rT_2=-\frac{MR^2}{2}|\alpha|,\\
T_1=m(g-a),\\
|\alpha|=\frac{a}{r}
\end{cases}$$, from which it follows that $a=\frac{m-3\mu_k M}{m+3M}g \approx 3.40 m/s^2.$

 

[TSny]

Your work looks correct to me.

 

[haruspex]

Me too, but I'd prefer not to use $|\alpha|$. Taking the usual convention of anticlockwise positive and positive to the right for the acceleration we can write
$Rf_d-rT_2=\frac{MR^2}{2}\alpha$
$a=r\alpha$
leading to a negative value for a.