Constraint Relations Homework: Acceleration of Block

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SUMMARY

The discussion focuses on deriving the relationship between the accelerations of a block and a loop in a pulley system, specifically using the angle theta (θ) and the acceleration of the loop (a). The key conclusion is that the acceleration of the ring (a_ring) is equal to the acceleration of the block (a_block) multiplied by the cosine of theta (cos θ). This relationship is established through the application of the Pythagorean theorem and differentiation, leading to the formula a_ring = a_block * cos θ.

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  • Knowledge of calculus, particularly differentiation.
  • Ability to interpret and analyze diagrams related to mechanics.
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  • Study the principles of kinematics in pulley systems.
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ritwik06
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Homework Statement


http://img510.imageshack.us/img510/5505/systemet4.jpg

What I wish to do is to relate the accelerations of the loop an the massive block. I know the angle theta at any instant. I also know that the acceleration of the loop on the fixed support is a. I have been given no other information except the figure and I have to write out the acceleration of the block.
My teacher says its a cos theta

But I cannot express the expression of the block in just two variable of a and theta. I also need the perpendicular distance of the fixed support from the pulley.

please help me!
 
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Referring to the diagram:

The Perpendicular distance from the pulley to the axis of the ring is say a constant L. The part
of the string that is being pulled (variable) is say y. The movement of the ring on the X-axis is also changing (variable) and we take it as x.

Now, using the hypotenuse theorem (!) :

x^2+ L^2 = y^2

what we want is now the instantaneous change of the variable x and y with respect to time and the way to find that is by differentiating:

2x\frac{dx}{dt} + 0 = 2y\frac{dy}{dt}

or \frac{dy}{dt} = \frac{x}{y} . \frac{dx}{dt}

From the diagram \frac{x}{y} = cos\theta

therefore . v_{ring} = v_{block} . cos\theta

We can say now that:

a_{ring} = a_{block} . cos\theta

Hope that solves it.
 

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