A 6 -lb collar C is attached to a spring and can slide on rod A B, which in turn can rotate in a

A 6 -lb collar C is attached to a spring and can slide on...

Key Concepts

Relative Motion in Rotating Frames

In a setup where a mass moves relative to a rotating frame, it is important to distinguish between its motion relative to the frame and its absolute motion. The velocity of the mass relative to the rod (or rotating frame) can be different from its inertial frame velocity due to the rotation. Analyzing the system in terms of relative velocities allows one to break down the total motion into rotational and translational parts, which is essential for applying energy and momentum conservation laws accurately.

Spring Potential Energy

A linear spring stores potential energy when it is either compressed or extended from its undeformed length. This stored energy, given by one half times the spring constant multiplied by the square of the displacement, is a vital component in energy conservation analyses. In dynamic systems involving a spring and a moving mass, changes in the spring’s deformation directly contribute to the work done on the system and the resulting redistribution of kinetic energy.

Moment of Inertia

The moment of inertia quantifies how the mass is distributed relative to a given axis and determines how difficult it is to change the rotational motion of a body. In problems involving rotation, both the intrinsic moment of inertia of the rod and the contribution from additional masses (such as the collar) must be considered to accurately calculate rotational kinetic energy and angular momentum. It plays a central role in relating angular velocity to kinetic energy in rotating systems.

Conservation of Angular Momentum

When forces acting on a system produce no external torque about a given axis, the system’s angular momentum remains constant. In a rotating assembly where the collar slides along the rod, the radial spring force acts toward or away from the axis and does not affect the angular momentum. This means that as the collar’s radial distance from the rotation axis changes, the angular velocity must adjust accordingly to conserve angular momentum. This principle is especially important in problems analogous to a figure skater pulling in her arms.

Conservation of Mechanical Energy

In systems where non?dissipative forces are present, the total mechanical energy (the sum of kinetic energy and potential energy) remains constant. For a rotating system that includes translational motion of a sliding mass and elastic energy stored in a spring, any change in the system’s kinetic energy is balanced by an opposite change in the spring’s potential energy. This concept enables one to relate the mechanical energies at two different configurations and hence solve for unknown velocities or angular speeds.

Transcript

00:02 For this problem, we'll draw a diagram of our spring in each position.

00:08 So in position 1, which is initially, the spring has length 26 inches.

00:14 This is from pythagoras.

00:17 In position 2, when the collar is closer to point a, the spring has length 12 .5 inches.

00:24 So first let's calculate delta.

00:25 Delta is the distance that the spring stretches in each case.

00:29 We know that the undeformed length of the spring is 10 inches.

00:35 So in position 1, delta is 26 inches minus 10 inches, which is 16 inches.

00:49 The potential energy in this position, v1, we'll call it, is a half k delta squared.

00:58 So that's the potential energy stored in the spring.

01:00 And that's half times the spring constant which is 15 pounds per inch times delta squared which is 16 inches squared hence we get the potential energy in position one to be 1 ,920 inch pounds and so we can convert this to 160 foot pounds so now doing the same for position 2, delta is equal to 12 .5 inches minus the undeformed length of 10 inches, and hence the spring is stretched 2 .5 inches.

02:06 The potential energy in position 2 is a half k delta squared.

02:13 So that's a half.

02:14 Again k is 15 pounds per inch delta is 2 .5 inches and we square that so we left with potential energy of the spring in position 2 .875 inch pounds and this is 3 .91 in foot pounds so we have the potential energy in each case.

03:02 So here we have a diagram of our rod ab in positions 1 and 2.

03:08 In position 1, the distance between a and the center of color c is 2 feet.

03:14 The system rotates with angular velocity omega 1, and c has v0 in the first position to be r1, omega 1, and component vr, of the velocity in the first position is zero.

03:33 In position two, the distance between a and c is r2.

03:38 The rod rotates with angular velocity omega 2, and c has components of its velocity, vr2, and v02 is equal to r2 times omega 2.

03:51 So from kinetics, since we know the moments of all forces about the shaft at point a are 0, this means that angular momentum about a in position 1 must equal the angular momentum about the same point a when the system is in position 2.

04:18 So let's write down what these momentum are.

04:22 So in position 1, the angular momentum is ir omega 1, where ir is the mass moment of inertia of the rod times omega -1.

04:37 1 plus the mass of color c times v0 times r1 that's the contribution in position 1 in position 2 the moment of inertia i r multiplied by omega c not omega c but omega 2 in position 2 plus the mass of c times v0 this is v0 in position 1 is v0 in position 2 times r2.

05:29 So the angular momentum is basically conserved for each position.

05:34 So we can write this as ir plus mcr1 squared.

05:43 We just replace v0, v0 1 times omega 1.

05:51 And this is equal to, we do the same on the right hand side.

05:54 This is ir plus mcr2 squared and the common factor is omega 2.

06:03 So we just replace v02.

06:11 And now we can apply our data.

06:15 So our given data tells us that the moment of inertia of the rod is equal to 0 .35 pound foot second squared.

06:35 The mass of color c, mc, is equal to its weight, which is 6 pounds over g.

06:45 32 .2 feet per square second.

06:50 R1 is equal to two feet.

06:57 R2 is equal to 7 .5 over 12, and that's also in feet.

07:07 The initial angular speed, omega 1, is 12 radiance per second.

07:15 So that's the angular speed in position 1.

07:18 So now if we go back to our equation above, we can start substituting our values in.

07:27 So what we get is that 0 .35 will suppress the units here, plus 6 over 32 .2 into 2 squared all times 12 radiance per second is equal to, in the right hand side, we have all these values again.

07:55 0 .35 plus 6 over 32 .2 into 7 .5 over 12 and that's squared multiplied by omega 2...

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