18-Nov-2016: Moment of Inertia and Frictional Torque

Apparatus: A disk and a frictionless cart.

Purpose: Find out the moment ot inertia of the disk and frictional torque. Calculate the time that the cart should take to slipe down a distance. Compare the theory time and experiment time.

Model: τc-τf=Iα

Process: 

    The "disk" is make up of two cylinder and a disk; therefore, in order to find out the moment of inertia of the entry "disk", we need to calculate their moment of inertia separately. First, we measure the diameter and the heigh of two cylinder and the the disk. We get:

    Then, we use the equation V=πh(d/2)^2 to find out the volume.

                                                    V1=39.3381 cm^3
                                                    V2=479.9128 cm^3
                                                    V3=40.1093 cm^3
                                                   
                                        Vtotal=V1+V2+V3=559.3602 cm^3

    In order to find out the mass of each part, we use  V/Vtotal=M/Mtotal. We can get the total mass from the mark on the disk. Mtotal=4.724 kg.

                                                M1=(V1/Vtotal)*Mtotal=0.3322 kg
                                                M2=(V2/Vtotal)*Mtotal=4.053 kg
                                                M3=(V3/Vtotal)*Mtotal=0.3387 kg

    Now, we can use calculate out the moment of inertia.
                                   
                                 cylinder: I1=1/2M1R1^2=4.0942*10^-5
                                               I2=1/2M2R2^2=2.0366*10^-2
                                      disk:  I3=1/2M3R3^2=4.0949*10^-5

                                              Itotal=I1+I2+I3=0.020448 kg*m^2

    Then, we need to find out the frictional torque. Before that, we need the angular deceleration. We spin the disk, and we measure the time it take to stop and the number of revolution. We get:

                                                      Δθ=9π, Δt=5.35 s, ωf=0 m/s

   We use:                                          ω0-αt=ωf  
                                                         ω0t-(αΔt^2)/2=Δθ

   By these two equation:                    (αΔt^2)/2=Δθ  
                                                         α=(2Δθ)/(t^2)=1.9757 rad/s

   We can get the frictional torque:      τf=Itotal*α=0.040399 N*m
   
    Then                                              TR3-τf=Iα
                                                          sinθMg-T=Ma
                                                          θ=49°
   
    By them, we get                            a=(sinθMgR3^2-τfR3)/(MR+I)=0.012931 m/s

    By                                                (at^2)/2=D, D=1 m
                                                         
    We get the theory time                  t=12.35 s

    Then, we set up the inclind, tie one end of a string on a cylinder and the other end of the string on the frictionless cart which is 0.502 kg. We adjust teh disk and the ramp in order to let the string be parallel to the ramp.

    The cart rolls down the inclined track for a distance of 1 meter, and we calculate how long it take for the cart to travel 1 meter from rest. We do it several time and take the average.
                                                                   t1=12.27 s
                                                                   t2=12.92 s
                                                                   t3=12.14 s
                                                                   taverage=12.44 s
    We compare the theory time and experiment time:
                                          theory time=12.35 s≈ 12.44 s =experiment time
                                                             There are 0.7% off.
    The frictionless cart is not actually frictionless. Therefore, the actual time may be longer than the theory time. It makes sense for our data.