Wheel pulled by cord on inner drum, 50 kg

Assumptions. The cord leaves the bottom of the inner drum (radius $r=0.060$ m) and pulls horizontally to the right; the wheel (outer radius $R=0.100$ m) rests on rough ground. Take $x$ to the right; the sense of $\alpha$ is stated in each step. The friction force $F$ acts at the contact point.

Data

$m = 50$ kg, $\bar k = 0.070$ m, $P = 200$ N, $\mu_s = 0.20$, $\mu_k = 0.15$

$$\bar I = m\bar k^2 = 50(0.070)^2 = 0.2450\ \text{kg m}^2, \qquad N = mg = 50\times9.81 = 490.50\ \text{N}$$

Step 1: assume rolling without slipping

Rolling to the right means the wheel turns clockwise, so $\bar a = \alpha R$ with $\alpha$ clockwise. The pull $P$ (below $G$) gives a counter-clockwise moment $Pr$ about $G$; friction $F$, taken to the left at the contact point, gives a clockwise moment $FR$:

$$\begin{aligned} P - F &= m\bar a\\ FR - Pr &= \bar I\alpha \end{aligned}$$

Eliminating $F$ with $\bar a=\alpha R$:

$$\alpha = \frac{P(R-r)}{m(\bar k^2 + R^2)} = \frac{200(0.100-0.060)}{50(0.0049+0.0100)} = 10.74\ \text{rad/s}^2\ \text{(clockwise)}, \qquad \bar a = \alpha R = 1.07\ \text{m/s}^2$$

Friction required: $F = P - m\bar a = 200 - 50(1.07) = 146.31$ N.

Maximum available: $F_{max} = \mu_s N = 0.20\times490.50 = 98.10$ N.

Since $F = 146.31\ \text{N} > F_{max}$, the wheel slips. The assumption is wrong, and kinetic friction acts: $F = \mu_k N = 0.15\times490.50 = 73.58$ N.

Step 2: sliding (kinetic friction)

$$\begin{aligned} \bar a &= \frac{P - F_k}{m} = \frac{200 - 73.58}{50} = 2.53\ \text{m/s}^2\ (\to)\\ \alpha &= \frac{Pr - F_kR}{\bar I} = \frac{200(0.060) - 73.58(0.100)}{0.2450} = 18.95\ \text{rad/s}^2\ \text{(counter-clockwise)} \end{aligned}$$

With $\alpha$ counter-clockwise, the contact point accelerates forward relative to the ground ($\bar a + \alpha R>0$), so it slips forward. This confirms that friction acts to the left.

Answer: $\bar a_G = 2.53$ m/s² to the right; $\alpha = 18.95$ rad/s² counter-clockwise.