A box and its contents have a total mass M. A string passes...

A box and its contents have a total mass $M$. A string passes through a hole in the box (Figure 9.57 ), and you pull on the string with a constant force $F$ (this is in outer space-there are no other forces acting).(a) Initially the speed of the box was $v_{i}$. After the box had moved a long distance $w,$ your hand had moved an additional distance $d$ (a total distance of $w+d$ ), because additional string of length $d$ came out of the box. What is now the speed $v_{f}$ of the box? (b) If we could have looked inside the box, we would have seen that the string was wound around a hub that turns on an axle with negligible friction, as shown in Figure $9.58 .$ Three masses, each of mass $m$, are attached to the hub at a distance $r$ from the axle. Initially the angular speed relative to the axle was $\omega_{i} .$ In terms of the given quantities, what is the final angular speed rclative to the axis, $\omega_{f} ?$

Key Concepts

Work-Energy Principle

This fundamental concept states that the work done by forces on an object results in a change in its kinetic energy. It underpins the analysis of systems where forces perform work over a distance, leading to an increase in the object's energy, whether the motion is translational, rotational, or both.

Translational Kinetic Energy

Translational kinetic energy is the energy associated with an object’s motion along a straight path. It is quantified as one-half of the mass multiplied by the square of the velocity. This concept is essential when analyzing objects that are accelerating or decelerating due to applied forces.

Rotational Kinetic Energy

Rotational kinetic energy pertains to the energy an object possesses due to its rotation about an axis. It is given by one-half of the moment of inertia times the square of the angular velocity. This energy component becomes significant in systems where parts of the object are rotating, such as wheels or hubs.

Moment of Inertia

The moment of inertia is a measure of an object's resistance to changes in its rotational motion about a specific axis. It depends on how the mass is distributed relative to that axis. A proper calculation of moment of inertia is crucial for understanding and predicting how a rotating system will behave under applied torques.

Relationship between Linear and Rotational Motion

A key concept is the connection between linear and angular quantities, often expressed as v = r ?, where v is the linear speed at the rim of a rotating object, r is the radius, and ? is the angular speed. This relationship is instrumental in converting energy and motion between translational and rotational forms.

Transcript

00:01 For this problem on the topic of angular momentum, we have a uniform horizontal disk of mass 100 kilograms and radius 5 .5 meter, turning without friction at 2 .5 revolutions per second on a vertical axis through its center, as we can see in the figure.

00:16 The angular speed of the disk is maintained as constant due to a drive motor at a, and while the disk turns a 1 .2kg mass on top of the disk slides outward in a radial slot.

00:31 The mass starts at the center of the disk at t is equal to zero and moves outward with constant speed 1 .25 centimeters per second relative to the disk.

00:40 It reaches the edge at 440 seconds.

00:44 Its motion is constrained by a brake at b to keep the radial speed constant.

00:50 Now, we want to find the torque as a function of time, the value of this torque at 440 seconds, the power that the drive motor must deliver as a function of time, the value of the power when the sliding mass is just reaching the end of the slot, the string tension as a function of time, the work done by the drive motor during the 440 seconds, the work done by the string, string brake on the sliding mass, and the total work done in the system consisting of the disk and the mass.

01:23 So firstly, the sliding coordinate or the radial coordinate of the sliding mass, r of t is equal to 0 .0 .0 .125 meters per second times t.

01:41 Its angular momentum, l, is equal to mr squared omega, which is 1 .2 kg times 2 .5 revolutions per second, we'll convert this to radiance, so we multiply this by 2 pi radiance per revolution times 0 .0 .125 meters per second squared times t squared.

02:23 And so the angular momentum as a function of time is 2 .95 times 10 to the minus 3 kg meter squared per second cubed multiplied by t squared.

02:43 Now the drive motor must supply a torque that is equal to the rate of change of this angular momentum.

02:50 So tor is dl d t, which is if we take this expression and find the derivative with respect to time, we get this to be 0 .00589 watts times t.

03:17 And so that's the torque that needs to be supplied by the drive motor.

03:24 For part b, we want to find the value of this torque at 440 seconds.

03:31 So we'll call this tor f, and this is 0 .00589 watts times the time of 440 seconds, which gives us the torque to be 2 .59 watts.

03:52 Newton meters.

04:00 For part c, we want to find the power that the drive motor delivers as a function of time.

04:07 So this power p is equal to the torque tor times the angular speed omega.

04:14 And so this is 0 .00589 watts times t times 5 pi radians per second.

04:35 So this gives us the power of 0 .095 watts per second times t.

04:56 For part d, we want to find the value for the power when the sliding mass is just reaching the end of the slot...

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