The 0.5-1 b ball is guided along the vertical circular path r=2 rc cosθusing the arm O A. If the

The 0.5-1 b ball is guided along the vertical circular path...

Key Concepts

Newton's Second Law in Rotational Contexts

Newton's second law states that the net force acting on a body is equal to the mass of the body multiplied by its acceleration (F = ma). When applied to rotational or curvilinear motion, this law is used in conjunction with the decomposed acceleration components (radial and tangential) to determine the required physical forces. This framework is essential in problems where certain forces, like the force provided by an arm or a guide, must be computed from the known acceleration of a mass during rotation.

Dynamics of Circular Motion

Dynamics of circular motion involves applying Newton's second law to situations where an object undergoes rotation or is constrained to move along a circular path. This includes understanding centripetal (radial) acceleration, which is directed towards the center of the circle, and any tangential acceleration that may arise from changes in the speed of the object. Analyzing the net forces acting on the object along the radial and tangential directions allows one to determine the forces required to maintain circular motion.

Constraint Forces in Guided Motion

Constraint forces are those forces that arise from the geometric or physical constraints imposed on an object's motion, ensuring that it follows a particular path. In the context of a particle being guided along a specified path—such as a ball moving along a circular arc—the constraint force is the reactive force exerted by the guiding mechanism (the arm). This force ensures that the object stays on the prescribed path, and its calculation involves considering both the centripetal (radial) components and any contributions from tangential acceleration.

Kinematics in Polar Coordinates

This concept involves describing the motion of a particle in terms of radial and angular (or tangential) components. In problems with motion along a curved or circular path, the position is expressed in terms of a radius (r) and an angle (?). The velocity and acceleration of the particle are derived by differentiating these quantities with respect to time, leading to expressions for radial acceleration (including the centripetal term) and tangential acceleration. This framework is crucial for analyzing motions where the path is defined by a polar equation.

Angular Kinematics

Angular kinematics deals with the motion of a body as it rotates, characterized by angular velocity and angular acceleration. These quantities describe how fast the angle is changing and how the rate of change itself is varying over time. In problems involving rotation, the angular variables provide the basis for determining how the position, velocity, and acceleration components evolve, and they are essential when the path of a particle is defined in terms of an angle.

Transcript

0:00 Hi everyone.

00:01 So what we have is a .5 pound ball that it's going along a vertical circular path given by r equals 2 times r c coast theta.

00:12 And we know that the rod or the angular arm, the arm has an angular velocity of 0 .1 radiance per second and that it's accelerating at 0 .8 radiance per second squared.

00:23 What we're asked to solve for is the force of the arm on the ball when theta is equal to 30 degrees, that the rc is equal to 0 .4 feet.

00:34 And the way that we're going to be doing this is using our knowledge of tangential and angular motion and using free body diagrams to develop equations of motions to solve with this force.

00:47 So let's go ahead and get started.

00:49 First thing we're going to do is we're going to say that r is equal to 2 times rc close data.

00:56 And since we know rc is just 0 .4 feet, we can say that r is equal to 0 .8 cost theta.

01:03 Therefore, r dot is equal to negative 0 .8 sine theta times theta dot.

01:12 Fix this data here.

01:15 And we also know that our double dot is equal to minus 0 .8 cost theta times theta dot squared minus 0 .8 sine theta times theta double dot.

01:30 So what we can do is knowing our knowledge that at theta equal to 30 degrees, we know that theta dot is 0 .4 radians per second, and that theta double dot is equal to 0 .8 radiance per second squared.

01:49 We can go ahead and solve for our r, r dot, and r double dot.

01:53 So by plugging all the information we know, we know that r is just going to be equal to 0 .69 feet.

01:58 We know our dot is going to be equal to minus 0 .16 feet per second...

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