Growing up I’ve had many memorable and important experiences. A lot of these situations have led me to become a better person. I can say that I have had my share of bad experiences as well, but all have led up to making me the person I am today. One experience that sticks out among them all is one a little over nine years ago. when I learned a valuable lesson. The most memorable childhood experience I have recalled is when I was nine years old. I grew up with my mom, dad, two sisters and a brother. Being the oldest, I always played as a babysitter for my parents

. One day my parents were getting ready to take my brother and baby sister to the doctor. They went through their precautionary measures before they left like ; locking all the doors, windows, leaving their numbers on the fridge, and so on. before walking out the door my dad told me to do my homework and watch my sister. As they left I decided to disregard everything said and not do my homework or even keep an eye on my sister. I started playing Playstation, I got the new Ratchet and Clank game a week earlier and didn't want to waste any time in beating it. As the day progressed and I was getting further in the game, I started thinking about when my parents were going to get home. So I paused the game for a little to call them. Getting a hold of them was pretty hard considering my parents never answer their phones, but I eventually got through to my mom. During our conversation my mom asked if I did my homework, tensing up and scared to say no, I replied yes, then she asked about my sister. I had no clue where she could be but I again I replied with a hesitant," y-y-yes mom she's in her room sleeping".

Already at lie number two, I swiftly rushed to my sisters room. Slamming the door open in urgency, I quickly scanned the room for any sign of my sister. nowhere to be found, I yelled out "CELINE!" multiple times. Getting no response, I ran down stairs hastily in search for my sister, thinking...

...Sadie D. Hood Lab 8: Moment of Inertia Partner: Florence Doval Due 16 November 2011 Aims: To use a centripetal force apparatus to calculate the moment of inertia of rotating weights, using theories derived from ideas of energy transfer (Im = MR2 (g/2h)(t2-t02)) and point mass appoximation (m1r12 + m2r22). Set Up
Procedure First we measured the weights of two masses and wingnuts that secure them. Then we placed one of the masses on the very end of a horizontal rod on the centripetal force apparatus, 0.162 m away from the centre of the rod, and the other mass 0.115 m away from the centre of the rod. Then we attached a 0.2 kg mass to the bottom of a string and wound the string around the vertical shaft of the apparatus, so that the bottom of the weight rose to the bottom edge of the tabletop the apparatus was on. We measured the distance from the bottom of the weight to the floor, and then let the weight fall to the floor, and measured the time it took to do so. We repeated this measurement, using the same initial height, four more times. Then we took the masses and the wingnuts off the horizontal rod and let the 0.2 kg mass fall in the same way as before, five times. Then we replaced the masses and the wingnuts, but put them both on the edges of the horizontal rod, and repeated the same falling mass measurements five times. We moved the masses in towards the centre of the rod and continued to repeat the falling mass measurements. We moved the...

...Moment of Inertia
1. Abstract
The goal of this study is to understand the transfer of potential energy to kinetic energy of rotation and kinetic energy of translation. The moment of inertia of the cross arm my group measured with the conservation of energy equation is: 0.01044 kg/m2 (with the mass of 15g), 0.01055 kg/m2 (with the mass of 30g), which is kind of similar to the standard magnitude of the moment of inertia of the cross arm: 0.0095 kg/m2 (Gotten by measuring the radius and the mass of the cross arm and use the definition equation of moment inertia). And we also get the moment of inertia for disk: 0.00604 kg/m2, and the ring: 0.00494 kg/m2.
2. Introduction
For that experiment, we use the conservation of energy equation to find the moment of inertia of the cross arm, and then use the definition equation of the moment of inertia to get the exact magnitude for the cross arm, the disk and the ring.
For the first step, finding the moment of inertia of cross arm, we need the conservation of energy equation for the transferring of the energy from the potential energy to the sum of the kinetic energy for both the mass strikes and the cross arm, like equation 1:
Since the mass is falling with uniform acceleration, its final velocity, v, after having fallen through height h can be found by using equation 2:
And the angular velocity of the...

...[pic] The flywheel of an engine has moment of inertia 2.5 kg•m2 about its rotation axis. What constant torque is required to bring it up to an angular speed of 400 rev/min in 8s, starting from at rest?
[pic] A solid, uniform cylinder with mass 8.25kg and diameter 15cm is spinning at 200 rpm on a thin, frictionless axle that stop the cylinder axis. You design a simple friction brake to stop the cylinder by pressing the brake against the outer rim with a normal force. The coefficient of kinetic friction between the brake and rim is 0.333. What must be the applied normal force to bring the cylinder to rest after it has turned through 5.25 rev?
[pic] A 2.2kg hoop 1.2m in diameter is rolling to the right without slipping on a horizontal floor at a steady 3 rad/s. (a) how fast is its center moving? (b) What is the total kinetic energy of the hip? (c) Find the velocity vector of each of the following points as viewed by a person at rest on the ground: i) the highest point on the hoop; ii) the lowest point on the hoop; iii) the point on the right side of the hoop, midway between the hoop and the bottom. (d) Find the velocity vector for each points in part c, except as viewed by someone moving along with same velocity as the hoop.
[pic] A solid ball is released from res and slides down a hillside that slopes downward at 65° from the horizontal. (a) What minimum value must the coefficient of static friction between the hill and ball surfaces have for no...

...Title: Mass Moment of Inertia
Objective:
To determine mass moment of inertia of a part using experimental method.
Theory:
If a part has been designed and built, its mass moment of inertia can be determined approximately by a simple experiment. This requires that the part be swung about any axis (other than one that passes through its CG) parallel to that about which the moment is sought and its period of pendular oscillation measured. Figure 1 shows a part of connecting rod suspended on a knife-edge pivot at ZZ and rotated through a small angle θ.
Its weight force W acts as its CG has a component W sin θ perpendicular to the radius r from the pivot to the CG.
From rotational form of Newton’s equation:
TZZ=IZZ∝
Substituting equivalent expressions for TZZ and ∝;
-Wsin θr=IZZd2θdt2
Where the negative sign is used because the torque is in the opposite direction to angle a.
For small values of θ, sinθ=θ, approximately, so:
-Wθr=IZZd2θdt2
d2θdt2=-WrIZZθ
Equation above is a second order differential equation with constant coefficients that has the well-known solution:
θ=CsinWrIZZt+DcosWrIZZt
The constants of integration C and D can be found from the initial conditions defined at the instant the part of released and allowed to swing.
At: t=0, θ=θmax, ω=dθdt=0;then:C=0, D=θmax
And:
θ=θmaxcosWrIZZt
Equation above defines the part’s motion as a cosine wave that completes a full cycle of period τ sec when...

...THE MOMENTS OF A RANDOM VARIABLE
Definition: Let X be a rv with the range space Rx and let c be any known constant. Then the kth moment of X about the constant c is defined as
Mk (X) = E[ (X c)k ]. (12)
In the field of statistics only 2 values of c are of interest: c = 0 and c = . Moments about c = 0 are called origin moments and are denoted by k, i.e., k = E(Xk ), where c = 0 has been inserted into equation (12). Moments about the population mean, , are called central moments and are denoted by k, i.e, k = E[ (X )k ], where c = has been inserted into (12).
STATISTICAL INTERPRETATION OF MOMENTS
By definition of the kth origin moment, we have:
k =
(1) Whether X is discrete or continuous, 1 = E(X) = , i.e., the 1st origin moment is simply the population mean (i.e., 1 measures central tendency).
(2) Since the population variance, 2, is the weighted average of
deviations from the mean squared over all elements of Rx, then 2 =
E[(X )2] = 2. Therefore, the 2nd central moment, 2 = 2, is a measure of dispersion (or variation, or spread) of the population. Further, the 2nd central moment can be expressed in terms of origin moments using the binomial expansion of (X )2, as shown below.
2 = E[ (X...

...Laboratory VII: Rotational Dynamics
Problem #1: Moment of Inertia of a Complex System
John Greavu
May 8, 2013
Physics 1301W, Professor: Evan Frodermann, TA: Mark Pepin
Abstract
The moment of inertia of a complex system was determined through two different approaches. A string was tied to and tightly wound around a horizontal disc and then strung over a vertical pulley where the other end was then tied to a hanging weight. Underneath and attached through the center of the disc were a spool and shaft (all supported by a stationary stand). A ring was also centered on top of the disc. The shaft, spool, disc, and ring were free to rotate together. The string was then allowed to unravel around the disc, which also allowed the hanging mass to descend. A video of this was taken, uploaded, and plotted in MotionLab where it was used to construct horizontal position and horizontal velocity vs. time graphs. It was predicted that the moment of inertia of the system was equal to the sum of the individual moment of inertias (for each object in rotation). After analyzing the data and the subsequent graphs, the moment of inertia for the system was also determined as a function of the mass of the hanging weight, the radius of the disc, gravity, and the acceleration of the hanging weight. This function was then compared to the original prediction equation.
Introduction
“While examining the engine of your friend’s snow...

...Introduction
A bending moment is simply defined as “the algebraic sum of the moments of all the forces which induces bending of an element” (1). The aim of this assignment is to work out the bending moment in a simply supported beam when different concentrated loads are applied to it. A simply supported beam is a structure, usually with a straight profile supported at the ends, often pinned on one side and simply supported or on a roller on the other. There will be three series of loads applied to this beam & the findings will be recorded. The results will then be compared with the theoretical bending moment & the reasons for any variation explained.
The main reason for the experiment to be conducted is to examine, not only the accuracy of the testing equipment, but also the accuracy of bending moment calculations and diagrams compared to a real-world assessment. It will hopefully prove that “the bending moment at a cut section is equal to the algebraic sum of the moments acting to the left or right of the section”. (2)
After this introduction, there will be a little background information about this experiment and its apparatus, followed by a breakdown of the experimental procedure. Then, there will be the displayed results before a comparison with the theoretical results that have been calculated. Finally, while the conclusions are made, I will attempt to explain the...

...“Measurement of bending moment and
shear forces for structural analysis”
Azamat Omarov
ID201102658
1.Theory and background
1.1 Summary
That performed laboratory session on bending moments and shear forces requires good understanding and sufficient knowledge of axial forces. Bending is defined as a behavior of any structural element that undergoes the external load, which is applied perpendicularly to longitudinal axis. That experiment helps us to find the maximum load that can be applied to the beam with rectangular cross section. Moments are calculated by using statics theory, or multiplying perpendicularly directed load by the respective distance to the pivot point.
1.2 Objective
The main objective of that laboratory is to provide students with basic experience and thus, the comparison between calculated and measured values (software) should be demonstrated to show the ability to apply statics theory from applied mechanics module.
1.3 Theory
Shear forces
The shearing force at any section of a beam is the algebraic sum of the lateral components of the forces acting on either side of the section. F is the resultant reaction on the left of AA. As the beam is in equilibrium then resultant reaction on the right of AA must be downwards.
Figure1. Shear forces diagram
Equilibrium state
∑Fx=0N; ∑Fy=0N; ∑Mo=0N.m (1)
In our case we use AA as a reference point to calculate the bending...

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