#$&*
course phy 201
10/23 9:00 pm
`q001. A domino of mass .02 kg is moving at 140 cm / second. What is its kinetic energy?-KE = ½ m * v^2; 0.1 * 19600; KE = 1960 kg cm^2/s^2
If the kinetic energy of a domino of mass .02 kg is .5 Joules, how fast is it moving?
-0.5 Joules = 0.1 * v^2; 5 = v^2; v = roughly 2.24 m/s
An object of unknown mass is given a kinetic energy of 8 Joules, which causes it to move at 12 meters / second. What is its mass?
-8 Joules = ½ m * 144; 0.056 = ½ m 0.111 kg
All the above questions would be answered using a single definition. What is that definition?
-KE = ½ m * v^2, definition of kinetic energy
`q002. A domino of mass .02 kg is positioned on a rotating strap, 20 cm from the axis of rotation. How far does the domino move as the strap completes a full rotation, and how far does it move as the strap rotates through 220 degrees?
- radius, 20 cm; Distance travelled through one rotation equals circumference of that distance. Diameter: 40 cm; Circumference: 125.6 cm; Distance covered through one full rotation: 125.6 cm; 220 degrees is 61% of 360; distance covered at 220 degrees: 76.6 cm;
How fast is the domino moving if the strap is rotating at 220 degrees / second, and what is its kinetic energy at that speed?
-220 degrees/second, 76.6 cm covered in 220 degrees, average velocity: 76.6 cm/s; KE = ½ m * v^2; KE = 0.1 kg * 58.67 m^2/s^2; KE = 5.867 Joules
Through how many degrees per second would the strap have to be rotating in order for the domino to have 0.1 Joules of kinetic energy?
0.1 Joules = 0.1 * v^2; v = 1 m/s; 1 meter/s = roughly 79.6% of full rotation speed. 79.6% of 360 = roughly 286.5 degrees/second
These questions would be answered with the same definition used for the first question, in addition to what common knowledge from high school mathematics?
How to find circumference, as well as usage of percentiles.
`q003. A horse of mass 500 kg starts from rest and descends a steep bank, losing 50 000 Joules of gravitational potential energy. How fast would the horse be moving if all the lost potential energy was converted to kinetic energy?
-all energy converted into KE; KE = 50,000 Joules. 50,000 = 250 * v^2; 200 = v^2; 14.14 = v; speed = 14.14 m/s
If at the bottom of the bank the horse in the preceding is moving at 10 meters / second, how much work was done on it by nonconservative forces?
-4.14^2 = 17.14; 250 * 17.14 = 4285 Joules.
@&
The horse isn't moving at 4.14 m/s, so this doesn't work.
Find the KE at 10 m/s, and work from that result. You'll see how 4285 Joules is not correct. In fact, you'll find that the horse lost as much KE as it ended up with, i.e., half the 50 000 Joules.
The key error in your reasoning: vf^2 - v0^2 is very different from (vf - v0) ^ 2.
*@
What might be the nature of these nonconservative forces?
-One would automatically assume friction was involved.
@&
The horse is probably digging its hooves into the side of the hill, as opposed to sliding down the hill, though both are possible and both are almost certainly present. There is certainly some friction, but the reaction of the compressed soil to the action of the horse is more likely the main force in this situation.
*@
The first two questions were answered based on what definitions and what theorems?
Definition of KE, and the conservation of energy.
`q003. What is the kinetic energy of a rotating system whose moment of inertia is .012 kg m^2 when it is rotating at 4 radians / second?
-KE = ½ I * omega ^2; omega = angular velocity in radians/second; KE = 0.006 * 16 = 0.096; KE: 0.096 Joules;
What is the angular velocity of this rotating system if its kinetic energy is 0.6 Joules?
-0.6 Joules = 0.006 * omega^2; 100 = omega^2; 10 = omega. Angular velocity = 10 radians/second
What is the moment of inertia of a rotating system whose kinetic energy is 500 Joules when it is rotating at 40 radians / second?
-500 Joules = ½ I * 40^2; 500 Joules = ½ I * 1600; 0.3125 = ½ I; Moment of inertia = 0.625;
What definitions were required to answer these questions?
-KE = ½ I * omega^2; I = moment of inertia, omega = angular velocity in radians/second;
`q004. What is the moment of inertia of a styrofoam wheel in which 12 bolts, each of mass 100 grams, are inserted around a circle 50 cm in diameter and concentric with the axis of rotation, with another 8 bolts, each of mass 150 grams, inserted around a concentric circle 30 cm in diameter? Ignore the moment of inertia of the styrofoam itself.
-moment of inertia = 1/12 mass * L^2; calculate moment of inertia separately; 50 cm ring, 1200 grams; moment of inertia = 100 grams * 2500 cm; moment of inertia = 0.1 kg * 25 m^2; moment of inertia = 2.5 kg m^2; 30 cm ring, 1200 grams; moment of inertia = 100 grams * 900cm^2; moment of inertia = 0.1 kg * 9 m^2; moment of inertia = 0.9 kg m^2; inertia at 50 cm: 2.5 kg m^2, inertia at 30 cm: 0.9 kg m^2;
@&
1/12 M L^2 is the moment of inertia of a rod of mass M and length L.
There is no rod involved here.
The moment of inertia of a point mass is m r^2. The moment of inertia of a mass which is concentrated near a point is close to m r^2, and this is the case for a bolt.
*@
`q005. A long metal spring exerts forces of 500 Newtons, 400 Newtons, 250 Newtons, 100 Newtons, and 5 Newtons at respective lengths 20 meters, 18 meters, 16 meters, 13 meters and 10 meters.
What is its average tension between the 16 meter length and the 13 meter length?
-175 Newtons
How much work would be required to stretch the cord from the 13 meter length to the 16 meter length?
-525 Joules
If the tension is conservative, how much potential energy would be lost as the cord's length decreases from 16 meters to 13 meters?
-525 Joules
Assuming the tension to be conservative, how much potential energy is built on each of the four intervals, as the spring is stretched from its 10 meter length to its 20 meter length?
-2525 Joules
@&
The question asked for the work done on each interval, not the total for all intervals, and certainly not what you would get if you ignored everything but the first and last point.
Your result is an overestimate. Not a drastic overestimate but a significant one.
*@
If all that potential energy could be converted to the kinetic energy of a rotating system whose moment of inertia is 2 kg m^2, what would be the angular velocity of that system?
-2525 Joules = 1/12(2) * omega^2; 15119.8 Joules = omega^2; angular velocity = roughly 123 radians/second;
@&
Again 1/12 M L^2 is not relevant to this situation.
*@
What definitions are required to answer these questions?
-work = average force * displacement; work = change in kinetic energy; kinetic energy = 1/12(inertia) * omega^2, where omega is the angular velocity in radians/second;
`q006. How long would it take a freely falling object, starting from rest, to reach a speed of 300 meters / second, and how far would it travel during this time?
-acceleration: 9.8 m/s^2; v0 = 0 m/s; vf = 300 m/s; dv = 300 m/s; time to reach velocity: 30.6 seconds; vAve = 150 m/s; distance travelled: 4590 meters;
What would be the change in the gravitational potential energy of a 100 kg mass falling in the manner described here?
-4498200 Joules
If due to air resistance a 100 kg mass, falling from rest, requires 40 seconds to reach a speed of 300 meters / second, then what is your best estimate of the average force of air resistance?
-estimate: roughly 326 Joules
@&
326 Joules is an energy, not a force.
You need to detail your reasoning.
*@
@&
Very good on most questions.
You should rework the ones you missed, and submit them if you aren't completely sure of your solutions.
*@