Constraint Relations Homework: Acceleration of Block

[ritwik06]

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!

 

[ron_jay]

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.

 

[thomate1]

I would recommend approaching this problem using the principles of Newton's laws of motion. First, we need to identify all the forces acting on the block and the loop. The forces acting on the block are its weight (mg) and the tension in the string (T). The force acting on the loop is the normal force from the fixed support (N) and the tension in the string (T).

Next, we can write out the equations of motion for the block and the loop using Newton's second law (F=ma). For the block, we have:

ΣF = ma
T - mg = ma

For the loop, we have:

ΣF = ma
N - T = ma

Since we know that the acceleration of the loop on the fixed support is a, we can substitute this value into the equation for the loop:

N - T = ma
N - T = ma cosθ

We also know that the normal force (N) is equal to the weight of the block (mg) since there is no other external force acting on the loop. Substituting this into the equation, we get:

mg - T = ma cosθ

Now, we can solve for the acceleration of the block (a) by rearranging the equation:

a = (mg - T)/m cosθ

To find the tension in the string (T), we need to use the constraint relation between the block and the loop. This relation tells us that the length of the string is equal to the distance between the block and the pulley (d) plus the distance between the pulley and the fixed support (L). So, we have:

T = mg - ma cosθ

Now, we have an expression for the acceleration of the block in terms of the given variables (a and θ) and the distance between the block and the pulley (d). We can also use this expression to calculate the tension in the string.

In summary, as a scientist, I would suggest using Newton's laws of motion and the constraint relation to solve this problem. It is important to accurately identify all the forces acting on the block and the loop, and to use the given information to write out the equations of motion. From there, we can solve for the unknown variables and determine the acceleration of the block and the tension in the string.