Demonstration--Centripetal Acceleration vs. angular frequencyandCentripetal Force with a motor

 This lab shows the relationship of Centripetal Acceleration and angular frequency.

Our professor do a little lab and give us some different dates, then we should find the answer.

These are what we find. As the function, we know the mega=2pi/T, and angular acceleration=r*mega^2

At this case, we try to measure the Centripetal force with a motor.

first, we got the H=2m, h=1.15m, L=1.68m, R=0.98m, and it cost 2.07s per circle. Use the cos, we can find the theta. Lcostheta=H-h. the the theta=60 degree. Tantheta=mega^2*(Lsintheta+R)^2/(Lsintheta+R)*g.

then we get the mega=2.64rad/s. then we get the velocity is 6.43m/s.

If we use different time per circle, we will get different answer about velocity, theta, and mega.

Impulse-Momentum Activity

On this case we talk about a lab about Impulse and Momentum. I=F*T or I=M*V.

At this one, we use I=integralM*dv, then we can get the Impulse. 

at the second, we shuold use the same way to handle that.

As these four picture, we get the graph of force and velocity vs. times.

Projectile Motion

    On this case, we talk about the projectile motion with Newton's law.

The first picture shows us our lab and tools. in this step my partner should find the final velocity and the distance between the land point of ball and the table.

As the picture shows, we measure the theta which is 12 degree and the length is 0.79 meter, because the grav is 9.8m/s^2 and the mass of ball is 0.024kg, then the Fg is 0.2352N. then the F of ball is 0.2352*cos78, Because of F=ma, we know the acceleration is 2.04. then we know the t is 0.88s and the Vf is 1.6876m/s. (Allow the two function : Xf=Xi+Vi*t+1/2*a*t^2, and Xf=Xi+Vf*t-1/2*a*t^2.

As the second step, allow the same function, we know the time which the ball land, since the Vyi=0. The time is 0.4379s. and the Dx=0.65m, then the Vfinal =1.4814m/s.

at the second case, the every thing is same, so we need find the Distance.

Work-Kinetic Energy Theorem Activity

On this case, we use the logger pro to measure the Kinetic Energy. There are two Expts to do that.

At the first case, we put the cart to pull the spring, after it pull a position, release it from the initial position, then the logger pro will measure the graph about Force vs. Position. We note that, then we put different mass on the cart, and do it again. we note different date. As these pictures show.

the purple area is the work done area. As the graph show, the work done is should be same as work done area. the purple line is the total force and the black line means Kinetic Energy.

On the second case, this lab show the graph which is about Force vs. Stretch of the rubber band. The work done area is between the line which is force and the x-axis.

As this graph of lab show, there are some important dates on the picture. If we use the function about triangle(1/2*length*height), and the function of trapezoid(1/2*(top length+bottom length)*height), the area is the work done. Finally, we plus them all, we would get the total work done,

Magnetic Potential Energy Lab.

On this case, our lab is try to find the conservation of energy applies to this system. this lab tool try to make a environment that almost like the frictionless. 

As a frictionless environment, if we apply force to the cart, the force will be keep because there is no friction, then the acceleration is 0.

The first we know the cart's mass is 0.354kg, then we will measure the distance between two strong magnets, the angle and the force.

allow our dates, the angle has 0.5 degree error and the distance has 0.05cm error.

the we get a list:

1. r=0.031, F=0.242, theta=4

2. r=0.017, F=0.602, theta=10

3. r=0.012, F=0.898, theta=15

4. r=0.010, F=1.187, theta=20.

Then we used the logger pro to measure the date which used the function F=Ar^b. the we got the A=0.002357+/-0.0007603, and B=-1.349+/-0.07236.

Conservation Of Energy--Mass-Spring System

On this case, we will talk about the spring, mass and energy. The spring's is 0.044kg.
We use two different mass to measure the spring constant of spring.

Follow the date, when the spring hand 0.05kg, the distance is 0.05m and the force is 1.23N. because the hook's law, F=kx, we got the k is 16N/m.

On the second case, we let the spring hand 0.2kg. then the distance of spring is 0.2m and F is 3.68N, the we calculate that the K is 16.25N/m. it is almost closed.