Question 16.SP.10: 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 the coefficients of friction are μ_s = 0.20 and μ_k = 0.15, determine the acceleration of G and the angular acceleration of the wheel.
STRATEGY: Since you have forces acting on the wheel and are interested in accelerations, use Newton’s second law. Assume the wheel rolls without sliding and compare the friction force needed with the maximum possible friction force. If the force needed exceeds the force available, redo the problem assuming rotation and sliding.
MODELING: Choose the wheel as your system and model it as a rigid body. The acceleration of G is to the right and the angular acceleration is clockwise (Fig. 1). The free-body and kinetic diagrams for this system are shown in (Fig. 2).
ANALYSIS:
a. Assume Rolling without Sliding. In this case, you have
\bar{a}=r \alpha=(0.100 \mathrm{~m}) \alpha
The moment of inertia of the wheel is
\bar{I}=m \bar{k}^2=(50 \mathrm{~kg})(0.070 \mathrm{~m})^2=0.245 \mathrm{~kg} \cdot \mathrm{m}^2
Equations of Motion. Setting the system of external forces in your free-body diagram equal to the system of inertial terms in your kinetic diagram, you obtain
\begin{gathered}+\circlearrowright \Sigma M_C=\bar{I} \alpha + m \bar{a} d_{\perp}: \quad(200 \mathrm{~N})(0.040 \mathrm{~m})=\bar{I} \alpha + (m \bar{a})(0.100 \mathrm{~m}) \\8.00 \mathrm{~N} \cdot \mathrm{m}=\left(0.245 \mathrm{~kg} \cdot \mathrm{m}^2\right) \alpha + (50 \mathrm{~kg})(0.100 \mathrm{~m}) \alpha(0.100 \mathrm{~m}) \\\alpha=+10.74 \mathrm{rad} / \mathrm{s}^2 \\\bar{a}=r \alpha=(0.100 \mathrm{~m})\left(10.74 \mathrm{rad} / \mathrm{s}^2\right)=1.074 \mathrm{~m} / \mathrm{s}^2\end{gathered}
\begin{array}{ll}\stackrel{+}{\rightarrow} \Sigma F_x=m \bar{a}_x: \quad & F + 200 \mathrm{~N}=m \bar{a} \\& F + 200 \mathrm{~N}=(50 \mathrm{~kg})\left(1.074 \mathrm{~m} / \mathrm{s}^2\right) \\& F=-146.3 \mathrm{~N} \quad \quad \mathbf{F}=146.3 \mathrm{~N} \leftarrow\end{array}
\begin{aligned}+\uparrow \Sigma F_y &=m \bar{a}_y: \\N &-W=0 \quad N – W=m g=(50 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^2\right)=490.5 \mathrm{~N}\end{aligned}
\mathbf{N}=490.5 \mathrm{~N} \uparrow
Maximum Possible Friction Force.
F_{\max }=\mu_s N=0.20(490.5 \mathrm{~N})=98.1 \mathrm{~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 new free-body and kinetic diagrams (Fig. 3), where \bar{a} and α are independent and
F=F_k=\mu_k N=0.15(490.5 \mathrm{~N})=73.6 \mathrm{~N}
From the computation of part (a), you found that F is directed to the left. You can obtain and solve the following equations of motion as
\stackrel{+}{\rightarrow} \Sigma F_x=m \bar{a}_x: \quad 200 \mathrm{~N}-73.6 \mathrm{~N}=(50 \mathrm{~kg}) \bar{a}
\bar{a}=+2.53 \mathrm{~m} / \mathrm{s}^2 \quad \overline{\mathrm{a}}=2.53 \mathrm{~m} / \mathrm{s}^2 \rightarrow
\begin{aligned}+\circlearrowright \Sigma M_G &=\bar{I} \alpha: \\&(73.6 \mathrm{~N})(0.100 \mathrm{~m}) – (200 \mathrm{~N})(0.060 \mathrm{~m})=\left(0.245 \mathrm{~kg} \cdot \mathrm{m}^2\right) \alpha\end{aligned}
\alpha=-18.94 \mathrm{rad} / \mathrm{s}^2 \quad \alpha=18.94 \mathrm{rad} / \mathrm{s}^2\Lsh
REFLECT and THINK: The wheel has larger linear and angular accelerations under conditions of rotating while sliding than when rolling without sliding.
