Question 16.SP.9: A cord is wrapped around the inner drum of a wheel and pulle......
A cord is wrapped around the inner drum of a wheel and pulled horizontally with a force of 200 N. The wheel has a mass of 50 kg and a radius of gyration of 70 mm. Knowing that \mu_s=0.20 and \mu_k=0.15,, determine the acceleration of G and the angular acceleration of the wheel.
a. Assume Rolling without Sliding. In this case, we have
\bar{a}=r \alpha=(0.100 m ) \alpha
We can determine whether this assumption is justified by comparing the friction force obtained with the maximum available friction force. The moment of inertia of the wheel is
\bar{I}=m \bar{k^2}=(50\text{ kg} )(0.070 m )^2=0.245 \text{ kg} \cdot m ^2
Equations of Motion
+\downarrow \Sigma M_C=\Sigma\left(M_C\right)_{\text {eff }}: \quad(200 N )(0.040 m )=m\bar{a}(0.100 m )+\bar{I} \alpha
\begin{aligned}8.00 N \cdot m & =(50 \text{ kg} )(0.100 m ) \alpha(0.100 m )+\left(0.245 \text{ kg} \cdot m ^2\right) \alpha \\\alpha & =+10.74\text{ rad}/ s ^2 \\\bar{a} & =r \alpha=(0.100 m )\left(10.74\text{ rad}/ s ^2\right)=1.074 m / s ^2\end{aligned}
\begin{array}{lll}\stackrel{+}{\rightarrow} \Sigma F_x=\Sigma\left(F_x\right)_{\text {eff }}: & F+200 N =m \bar{a} & \\& F+200 N =(50 \text{ kg} )\left(1.074 m / s ^2\right) & \\& F=-146.3 N & F =146.3 N \leftarrow\end{array}
\begin{aligned}+\uparrow \Sigma F_y&=\Sigma\left(F_y\right)_{\text{eff}}: \\& N-W=0 \quad N-W=m \text{g}=(50 \text{ kg} )\left(9.81 m / s ^2\right)=490.5 N \\& &N =490.5 N \uparrow\end{aligned}
Maximum Available Friction Force
F_{\max }=\mu_s N=0.20(490.5 N )=98.1 N
Since F>F_{\max }, the assumed motion is impossible.
b. Rotating and Sliding. Since the wheel must rotate and slide at the same time, we draw a new diagram, where \overline{\mathbf{a}} and α are independent and where
F=F_k=\mu_k N=0.15(490.5 N )=73.6 N
From the computation of part a, it appears that F should be directed to the left. We write the following equations of motion:
\begin{aligned}+\downarrow \Sigma M_G&=\Sigma\left(M_G\right)_{\text {eff: }}: \\& (73.6 N )(0.100 m )-(200 N )(0.060 m )=\left(0.245 \text{ kg} \cdot m ^2\right) \alpha \\&\quad\quad\quad\quad\quad \alpha=-18.94\text{ rad} / s ^2 \quad \alpha=18.94 \text{ rad}/ s ^2 \uparrow\end{aligned}
