Dynamics rotating disk, solve for r and theta magnitudes

[lax1113]

Homework Statement

A screen clipping of the problem is here: http://img529.imageshack.us/img529/1738/dynamicsquestion2.jpg

We have a circular disk that roates about its center, 0, with a constant angular velocity [tex]\omega[/tex] ([tex]\omega[/tex] = [tex]\dot{\theta}[/tex]. The disk carries two spring-loaded plungers. The distance b, that each plunger protrudes from the rim of the disk varies according to b = b_o * sin (2[tex]\Pi[/tex]nt), where b_o is the max protrusion (n is a constant of oscillation, t is time).
Determine the maximum magnitudes of the -r and -[tex]\theta[/tex] components of the acceleration of the ends of the plunger during their motion.

Homework Equations

V_t = d[tex]\hat{u}[/tex]_r/dt

a_t = [tex]\omega[/tex]^2 (r_o + sin (2[tex]\Pi[/tex]nt)

The Attempt at a Solution

So with the At equation that is above, it is really obvious that the max r component of the acceleration will be when the trig portion is its max or 1. So this happens when t = n/4. We can substitute [tex]\dot{\theta}[/tex] in for omega, but I am not quite sure what it means by the maximum magnitude of the theta component.

Any hints would be appreciated, or if the direction I am headed in currently off.

thanks

 

[anathema]

Thank you for your question. The problem you have presented involves a rotating circular disk with two spring-loaded plungers attached to it. The motion of the plungers is dictated by the equation b = bo * sin(2\Pint), where bo is the maximum protrusion and n is a constant of oscillation. Your goal is to determine the maximum magnitudes of the r and theta components of the acceleration of the plungers during their motion.

To begin, let's consider the equation for the acceleration of a point on a rotating disk, which is given by at = \omega^2 (ro + sin(2\Pint)). Here, \omega is the angular velocity of the disk, ro is the distance from the center of the disk to the point, and t is time. As you correctly noted, the maximum r component of the acceleration will occur when the trigonometric function sin(2\Pint) is at its maximum value of 1. This occurs when t = n/4, where n is a whole number. Substituting this value into the equation, we can see that the maximum r component of the acceleration is given by armax = \omega^2 (ro + 1).

Now, let's consider the theta component of the acceleration. This component is perpendicular to the r component and is directed along the tangent to the circular path at the point where the plunger is located. The magnitude of this component can be found by considering the velocity vector, Vt = d\hat{u}r/dt, which is tangential to the circular path. The acceleration vector is then given by at = \omega * Vt. Since Vt is perpendicular to ro, the maximum theta component of the acceleration will occur when ro = 0, or when the plunger is at its maximum protrusion. This maximum theta component of the acceleration is given by atmax = \omega^2 * bo.

In summary, the maximum magnitudes of the r and theta components of the acceleration of the plungers are given by armax = \omega^2 (ro + 1) and atmax = \omega^2 * bo, respectively.

I hope this helps to clarify the problem for you. Let me know if you have any further questions.