What Forces Act on a Particle Confined in a Rotating Slot?

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The discussion focuses on analyzing the forces acting on a particle confined in a rotating slot, specifically at an angle of 30 degrees. The particle has a mass of 0.5 kg, with given angular acceleration and velocity values. The calculations involve determining the radial and tangential accelerations, leading to the normal force exerted by the slot on the particle. Despite attempts to solve the problem using cylindrical coordinates, discrepancies in the calculated normal force indicate a need to reassess the approach to focus on the horizontal motion of the particle. Ultimately, the conversation emphasizes the importance of accurately relating the particle's acceleration to the forces acting on it.
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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


\Sigma F_{\theta} = ma_{\theta} \\<br /> \Sigma F_{r} = ma_{r} \\<br /> a_{r} = \ddot{r} - r\dot{\theta}^{2} \\<br /> a_{\theta} = r\ddot{\theta} + 2\dot{r}\dot{\theta}

The Attempt at a Solution



My FBD.
https://lh4.googleusercontent.com/VAqoFLBBhCio8VVq62KJRvtmu9QsbRGa2Q9CcRHICXSSob2s-51TBrK10MZ0dZB6GQhlqn5YBxk=w1342-h547 [/B]
Establishing the position equation and taking time derivatives.
r=(0.5m)cos\theta \\<br /> \dot{r}=-(0.5m)sin(\theta)\dot{\theta} \\<br /> \ddot{r}=-(0.5m)[\ddot{\theta}sin\theta + \dot{\theta}^2cos\theta]<br />
Solving using the given values for theta, theta**, and theta*...
r|_{\theta=30}=(0.5m)cos(30)=0.43m \\<br /> \dot{r}|_{\theta=30,\dot{\theta}=2}=-(0.5m)sin(30)(2 rad/sec)=-0.5 m/sec \\<br /> \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<br />
Solving for ar yields
a_{r}=(-2.48m/sec^2)-(0.43m)(2rad/sec)^2=-4.20 m/sec^2
Sum of the forces in radial direction...
\Sigma F_{r}=Ncos(30)-mgcos(30)=ma_{r}
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. :(
 

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Can you explain why you're looking at the radial length r? Presumably the particle's motion is constrained to the x-direction and the arm slides (presumably without friction) against it. Perhaps you might consider finding an expression for the horizontal position of the particle in terms of θ, then see how its acceleration relates to the given parameters of θ for the instant in question.
 
gneill said:
Can you explain why you're looking at the radial length r?
I went right to looking at r since this question immediately follows a chapter on expressing motion in the cylindrical coordinate system (r,theta,z).

gneill said:
Perhaps you might consider finding an expression for the horizontal position of the particle in terms of θ, then see how its acceleration relates to the given parameters of θ for the instant in question.
I feel as though that would complicate the problem. Cylindrical coordinates simplify the expressions a little.
 
Haptic9504 said:
I went right to looking at r since this question immediately follows a chapter on expressing motion in the cylindrical coordinate system (r,theta,z).
But in this case you are looking for the forces acting on the particle, which is constrained to move horizontally. If you can determine its horizontal acceleration you can determine the net horizontal force acting on it. Then it's a bit of vector component work to sort out the rest.
 
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