- #1
Haptic9504
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Homework Statement
The particle has a mass of 0.5kg and is confined to move along the smooth horizontal slot due to the rotation of the arm OA. Determine the force of the rod on the particle and the normal force of the slot on the particle when θ = 30∘. The arm has an angular acceleration of θ¨ = 3rad/s2 and θ˙ = 2rad/s at this instant. Assume the particle contacts only one side of the slot at any instant.
Diagram is attached.
Homework Equations
[tex]\Sigma F_{\theta} = ma_{\theta} \\
\Sigma F_{r} = ma_{r} \\
a_{r} = \ddot{r} - r\dot{\theta}^{2} \\
a_{\theta} = r\ddot{\theta} + 2\dot{r}\dot{\theta}[/tex]
The Attempt at a Solution
My FBD.
https://lh4.googleusercontent.com/VAqoFLBBhCio8VVq62KJRvtmu9QsbRGa2Q9CcRHICXSSob2s-51TBrK10MZ0dZB6GQhlqn5YBxk=w1342-h547 [/B]
Establishing the position equation and taking time derivatives.
[tex]r=(0.5m)cos\theta \\
\dot{r}=-(0.5m)sin(\theta)\dot{\theta} \\
\ddot{r}=-(0.5m)[\ddot{\theta}sin\theta + \dot{\theta}^2cos\theta]
[/tex]
Solving using the given values for theta, theta**, and theta*...
[tex]r|_{\theta=30}=(0.5m)cos(30)=0.43m \\
\dot{r}|_{\theta=30,\dot{\theta}=2}=-(0.5m)sin(30)(2 rad/sec)=-0.5 m/sec \\
\ddot{r}|_{\theta=30,\dot{\theta}=2,\ddot{\theta}=3}=-(0.5m)[(3rad/sec^2)sin(3)+(2rad/sec)^2cos(30)]=-2.45m/sec^2
[/tex]
Solving for ar yields
[tex]a_{r}=(-2.48m/sec^2)-(0.43m)(2rad/sec)^2=-4.20 m/sec^2[/tex]
Sum of the forces in radial direction...
[tex]\Sigma F_{r}=Ncos(30)-mgcos(30)=ma_{r}[/tex]
Solving in terms of N and plugging in the variables gives me a value of N as 2.47N, when the accepted answer is 6.37N and based on the sum of forces in theta direction the value for F depends on N. I've done this problem over 3 times and still can't see where I am going wrong. :(