A bag is gently pushed off the top of a wall at A and swings in a vertical plane at the end of a

step 1 Hi everybody, thanks for joining me today. What we're going to be looking at is the concept of c

A bag is gently pushed off the top of a wall at A and swings...

Key Concepts

Conservation of Energy

This principle states that the total energy in an isolated system remains constant over time. In problems involving pendulum motion or swinging objects, gravitational potential energy is converted to kinetic energy as the object falls. This conversion is essential for relating the object's height (or angle) with its speed, which in turn affects the forces acting on it.

Centripetal Force

Centripetal force is the necessary net force that keeps an object moving in a circular path by constantly pulling it toward the center of the circle. It is given by the expression m*v^2/r, where m is mass, v is velocity, and r is the radius of the circular path. Understanding centripetal force is key to analyzing the forces in any circular motion, especially when determining tensions in a rope.

Tension in Circular Motion

In scenarios where an object is connected to a pivot by a rope or string and moves along a circular path, the tension not only provides the centripetal force but also counteracts the force of gravity. Analyzing the contributions to tension from both the weight component and the required centripetal force is crucial for predicting the conditions under which the rope might break.

Critical Load and Material Strength

Every material or component has a maximum load it can sustain before failure. In dynamics problems, this threshold often comes into play when forces exceed safe limits. By comparing the calculated tension in the moving system to the critical tension limit of the material, one can determine the conditions (such as a specific angle or speed) at which the material may fail.

Transcript

00:01 Hi everybody, thanks for joining me today.

00:03 What we're going to be looking at is the concept of circular motion with work and energy principles.

00:08 What we're given is we're given that a bag at a is gently pushed off that point and it is moving about with the radius of l and we're asked to find the angle that which the rope breaks and we know that the rope is going to break when the tension in the rope is equal to two times the weight.

00:30 First thing we're going to do is we're actually going to go ahead and draw a free body diagram.

00:33 So here is that mass.

00:36 We have a weight force acting down on it.

00:38 And we also have this tension force acting towards the center rotation.

00:44 As we draw our coordinates, we see that the normal force, the normal direction is directing the line with tension.

00:50 And we also have this tangential direction.

00:52 So as we resolve our forces in the normal tangential directions, we find that there is a component of the weight force that is acting in the direction and by using the angle theta we resolved that component to be w sine theta so given the convention we have with our coordinate system we can say that tension minus w sine theta is equal to m a and in the circular motion nomenclature the acceleration is equal to v squared over l so we can also say this is equal to mass times v squared over l we're going to come back to this later, and now we're actually going to move ahead into our kinetic energy arguments.

01:34 So obviously we know that the kinetic energy at the beginning is going to be zero because the bag is starting from rest.

01:40 So we see that t1 is equal to zero.

01:43 We know that t2 is going to be some kinetic energy depending on the final velocity.

01:48 That's going to be 1 .5 mv squared.

01:51 And we know that the work done from 1 to 2 is going to be due to the weight force that is active.

01:59 On the ball shown here, but remember it's only going to be acting over a component of the distance because gravity only acts in the vertical direction.

02:09 So this is going to be simply mgl sine theta, and l -sign theta is the vertical direction.

02:16 And you can see that here.

02:18 If we're starting from here, this is one, this is two, this is the radius l, and this is our theta.

02:26 You can see that this vertical component is l -sign theta.

02:29 All right, so we know that the changing kinetic energy is due to the work.

02:37 So we can say that t1 plus u1 to 2 is equal to t2.

02:42 We previously said that t1 is zero because the bag is starting from rest.

02:49 We know our expression for u1 -2 is mgl sine theta.

02:53 So we can say that mgl -s -s -theta is equal to one -half and a half and a new -two.

02:57 M bf squared...

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