A heavy particle is suspended by a 1 ·5 m long string. It is given a horizontal velocity of √(5

A heavy particle is suspended by a 1 ·5 m long string. It...

Key Concepts

Conservation of Mechanical Energy

This principle asserts that in the absence of non-conservative forces (like friction), the sum of kinetic and potential energy in a system remains constant. It is fundamental in analyzing systems where energy is exchanged between kinetic energy (due to motion) and potential energy (due to position in a gravitational field), such as in the motion of a pendulum or projectile motion.

Circular Motion and Centripetal Force

In any scenario where an object moves along a curved path, a centripetal force is required to keep it moving in that circle. For a pendulum, this force is provided by the tension in the string (along with contributions from gravity) while it is in circular motion. Understanding centripetal acceleration is vital when analyzing forces at various points along the object's path.

Critical Condition for Loss of Tension

This concept involves determining the point at which the force maintaining circular motion, such as the tension in a pendulum’s string, becomes zero. When the tension falls to zero, the object is no longer constrained to follow the circular path, marking a transition in the nature of its motion. Identifying this condition requires balancing the centripetal force requirement with the gravitational force components.

Projectile Motion Analysis

Once an object loses the constraint that forces it to follow a curved path (like when a pendulum's string goes slack), its subsequent motion can be analyzed as projectile motion. This involves decomposing its velocity into horizontal and vertical components and using kinematic equations to predict its trajectory, maximum height, and range under the influence of gravity.

Transcript

00:01 Okay, for this problem, we're considering a heavy particle that is suspended from a string, so that looks something like this, where a particle has some mass m, and they tell us that the length of the string is 1 .5 meters.

00:24 And we give that particle some horizontal velocity, v .0, where v .0 is equal to the square root of, of 57 meters per second.

00:40 We are additionally told to consider g equal to 10 meters per second squared.

00:47 And we want to know firstly at what angle will the line go slack.

00:55 This is the point where tension and the rope is equal to zero.

01:00 We want to know in part b, what is the velocity at that point, and in part c, we want to know what is the maximum height that the particle reaches.

01:15 So sometime later, our picture is going to look different.

01:19 So let's start by drawing what the new picture looks like.

01:24 So it tells us to consider theta from the vertical, and i'm going to assume that the particle makes it up here somewhere, but doesn't make it all the way around.

01:41 And you can check this using conservation of energy.

01:45 So i did that quickly.

01:46 I'm not going to go over it here.

01:48 But if you use conservation of energy, you'll see that the maximum height that this particle can reach, given that initial velocity is not enough to go all the way around.

02:01 So that means somewhere up here, the line is going to go slack.

02:06 Our angle theta that we're interested in knowing is what i just labeled on the graph there.

02:12 We also are interested in what the velocity is at this point and what is the height.

02:27 So to start, we need to in some way involve that angle theta in order to try and figure out what theta is equal to.

02:41 And so my first thought was either conservation of energy or newton's second law for rotational motion.

02:51 So this particle is going around in a circle, and we know that the sum of the forces in the radial direction should be equal to m v squared over r.

03:04 So i'm actually going to start with this equation in order to figure out what the maximum height is.

03:13 We could also look at conservation of energy as a starting point.

03:18 But i'm going to choose this as my starting point.

03:22 And so in order to do this, we of course want to draw a free body diagram.

03:26 So if this is our particle, we already know that if the string is slack, then the tension force is zero.

03:33 So the only force acting on this guy is mg downward, but it is moving in circular motion.

03:47 And so we need to figure out what the radial component of the gravitational force is going to be equal to.

03:58 Because we know, let me redraw that, because we know that this angle is theta, that tells us that this angle is also theta, which tells us that this angle is theta.

04:12 And so when we're considering the component of gravity that is in the radial direction, it's going to be the mg cosine theta component.

04:23 So now that we figured that out, that is the left -hand side of the equation here, and all of that has to be equal to mb squared over r.

04:38 We can of course cancel m and i'm going to replace r with l because that's what i've labeled the length of the string, which would correspond to the radius of the circle.

04:54 And we have a problem here, both theta and v are unknowns.

05:00 There are things that we're trying to find...

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