Constraint Relations Homework: Acceleration of Block

AI Thread Summary
The discussion focuses on relating the accelerations of a loop and a block in a physics problem involving a pulley system. The acceleration of the block is expressed as a function of the angle theta and the known acceleration of the loop, with the teacher stating it is a cos theta. The user derives the relationship using the Pythagorean theorem and differentiates to find the instantaneous changes in position. The final conclusion is that the acceleration of the ring is equal to the acceleration of the block multiplied by cos theta. This relationship is crucial for solving the homework problem effectively.
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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