The 0.5-lb particle is guided along the circular path using the slotted arm guide. If the arm has an angular velocity θ= 4 rad/s and an angular acceleration θ= 8 rad/s^2 at the instant θ =30° determine the force of the guide on the particle. Motion occurs in the horizontal plane.

Question

The 0.5-lb particle is guided along the circular path using the slotted arm guide. If the arm has an angular velocity \dot{ heta}= 4 rad/s and an angular acceleration \ddot{ heta}= 8 rad/{s}^{2} at the instant heta =30° determine the force of the guide on the particle. Motion occurs in the horizontal plane.

Accepted Answer

[latex]r=2(0.5 \cos \theta)=1 \cos \theta\[/latex]
[latex]\dot{r}=-\sin \theta \dot{\theta}\[/latex]
[latex]\bar{r}=-\cos \theta \dot{\theta}^{2}-\sin \theta \ddot{\theta}\[/latex]
At [latex]\theta=30^{\circ}, \dot{\theta}=4 \mathrm{rad} / \mathrm{s}[/latex] and [latex]\ddot{\theta}=8 \mathrm{rad} / \mathrm{s}^{2}\[/latex]
r=1 [latex]\cos 30^{\circ}=0.8660 \mathrm{ft}\[/latex]
[latex]\dot{r}=-\sin 30^{\circ}(4)=-2 \mathrm{ft} / \mathrm{s}\[/latex]
[latex]\ddot{r}=-\cos 30^{\circ}(4)^{2}-\sin 30^{\circ}(8)=-17.856 \mathrm{ft} / \mathrm{s}^{2}\[/latex]
[latex]a_{r}=\ddot{r}-\dot{r} \hat{\theta}^{2}=-17.856-0.8660(4)^{2}=-31.713 \mathrm{ft} / \mathrm{s}^{2}\[/latex]
[latex]a_{\theta}=\dot{r} \dot{\theta}+2 \dot{r} \dot{\theta}=0.8660(8)+2(-2)(4)=-9.072 \mathrm{ft} / \mathrm{s}^{2}\[/latex]
[latex]\nearrow +\Sigma F_{r}=m a_{r} ; \quad-N \cos 30^{\circ}=\frac{0.5}{32.2}(-31.713) \quad N=0.5686 \mathrm{lb}\[/latex]
[latex]+\nwarrow \Sigma F_{\theta}=m a_{\theta} ; \quad F-0.5686\sin30^{\circ}=\frac{0.5}{32.2}(-9.072)\[/latex]
F=0.143 lb

Question : The 0.5-lb particle is guided along the circular path using ...