A physics student of mass 43.0 kg is standing at the edge of the flat roof of a building, 12.0

A physics student of mass 43.0 kg is standing at the edge...

A physics student of mass $43.0 \mathrm{~kg}$ is standing at the edge of the flat roof of a building, $12.0 \mathrm{~m}$ above the sidewalk. An unfriendly $\operatorname{dog}$ is running across the roof toward her. Next to her is a large wheel mounted on a horizontal axle at its center. The wheel, used to lift objects from the ground to the roof, has a light crank attached to it and a light rope wrapped around it; the free end of the rope hangs over the edge of the roof. The student grabs the end of the rope and steps off the roof. If the wheel has radius $0.300 \mathrm{~m}$ and a moment of inertia of $9.60 \mathrm{~kg} \cdot \mathrm{m}^{2}$ for rotation about the axle, how long does it take her to reach the sidewalk, and how fast will she be moving just before she lands? Ignore friction in the axle.

Key Concepts

Conservation of Mechanical Energy

This concept involves the idea that in an isolated system, the total mechanical energy (the sum of kinetic and potential energies) remains constant if only conservative forces are doing work. In problems where gravitational energy is converted into kinetic energy (both translational and rotational), conservation of energy provides a powerful method to relate the initial and final states of the system.

Rotational Kinetic Energy

Rotational kinetic energy is the energy an object possesses due to its rotation and is given by one half the moment of inertia times the square of its angular velocity. In systems that couple translational and rotational motion, it is important to account for both translational kinetic energy and rotational kinetic energy when applying energy conservation.

Moment of Inertia

The moment of inertia is a measure of an object's resistance to changes in its rotational motion, analogous to mass in linear motion. It plays a crucial role in determining how energy is partitioned between translational and rotational forms in systems involving rotation, and it factors into the calculations of angular acceleration when torque is applied.

Relationship Between Linear and Angular Quantities

In rotational dynamics, there is a direct correspondence between linear and angular variables, typically expressed through the relation v = r?, where v is linear speed, r is the radius, and ? is the angular speed. This relationship is key for linking the translational and rotational dynamics in problems where an object unwinds or rolls without slipping.

Kinematics Under Constant Acceleration

Kinematics under constant acceleration involves equations that relate displacement, velocity, acceleration, and time when the acceleration is uniform. This concept is fundamental in determining quantities like time of fall or final speed when an object moves under a constant net force, and is often applied alongside energy methods in dynamics problems.

Transcript

00:01 In this question, a woman has jumped off of a 12 -meter roof holding onto a string that is attached to a wheel on an axle.

00:12 We're given that the axle's moment of inertia is 9 .6 kilograms per meter squared.

00:20 And that the woman's mass is 43 kilograms.

00:24 And the radius of the wheel with the rope around it is 0 .3 meters.

00:32 Given all that, we're asked to find how long it will take her to reach the bottom of the 12 -meter building and how fast we'll be going when she gets there.

00:40 So all of this fits very nicely into our energy equation that we've been using to solve problems like this.

00:49 E is equal to mgh, the energy from the woman falling.

00:53 The energy is going to be going into her, her kinetic energy, one -half mv squared, and the kinetic energy of the spinning wheel, one -half moment of inertia, times the angular velocity squared.

01:13 So we have almost all of these parts already to solve.

01:18 The only thing we're actually missing is what we want to solve this problem, which is the v.

01:23 That's our second question but it'll very easily help us find the t so focusing on that first seems like the good move first of all i'd like to change this into like terms so this be one -half mv squared plus one -half inertia which we have and then the angular velocity can be represented as v over r and of course that's squared so this will be two okay and now we can just pull out the v to the front equals v squared over two times m plus i over r squared and now all we need to do is move all this to the other side of the equation so at the end of that we get v squared equals 2mgh over m plus i over r squared...

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