5-Nov.-14: Moment of Inertia

PURPOSE:

To verify the sum-of torques equation, Στ=Iα, by finding the moment of inertia of a certain apparatus and directly applying the equation to predict how long it would take for a cart tied to our apparatus to accelerate down a 1-meter ramp.



APPARATUS:

  



  




Equipment -

Large metal disk on a central shaft, m = 4.840-kg
Calipers to measure the dimensions of the apparatus
Camera
1-meter ramp
String
Stopwatch
Rolling Cart, m = 500-g
Tape and marker (or any means of marking a specific point on the disk)


EXPERIMENT:

1. Using calipers and a ruler, we measured the apparatus' dimensions, which included:
   the radius of the cylinder, R1
   the radius of the disk, R2
   the width of the disk, w
   the distance from the surface of the disk to one end of the cylinder, x
   
   
   
2. We then calculated the moment of inertia for the entire apparatus.
   I = 0.021-kg*m^2
   
   
   
3. Using a piece of tape we marked a point at the edge of the disk.
   (red-dotted piece of tape on apparatus picture)
   We gave the disk an initial torque to get the apparatus spinning and measured the deceleration of the disk by recording the disk as it slowed to a stop.
4. We gave the disk an initial torque to get the apparatus spinning and measured the deceleration of the disk by recording the disk as it slowed to a stop.
5. To calculate the acceleration of the disk, we used logger pro to track the position of the red-dotted tape frame by frame. Tracking this specific point resulted in a graph (not included*) which gave us the acceleration tangential to the edge of the disk.
   a = -2.385-cm/s^2 = -0.02385-m/s^2
   
   We converted the tangential acceleration into rotational acceleration using the following equation which expresses the relationship between the two accelerations: a = r*αUsing r = R2 = 0.1002-mα = -0.238-rad/s^2
   
   *Faulty flash-drive
6. The apparatus spun to stopped due to a frictional torque. We were able to calculate this frictional torque with our calculated value for I and α
   
   Στ=IαΣτ=(0.021-kg*m^2)(0.238-rad/s^2)τf = 5.00*10^-3
7. We then attached a 500-gram cart to our apparatus with a string, with one end of the string tied to the cart and the other end tied and wound around the small cylinder part of the apparatus (R1). We also set secured a ramp near the apparatus at an angle of 40°(θ).
   
   
8. We derived a new expression for acceleration for an apparatus which now has a tied mass to it.
   (Note: In the calculations below, r_s is equal to the previously defined R1 and m is the mass of the cart.)
   
   
   
   We calculated the new acceleration to be a = 0.0326-m/s^2
9. We were able to use a kinematics equation to find a value for the time it would take for the cart to move down the ramp using our new found value for acceleration.
   
   
   
   t = 7.83-s is our prediction value.
10. We then released the cart on the ramp and let it roll toward the ground while simultaneously timing how long it took for the cart to reach the ground.
    We did this three times so that we could average the times, thus obtaining a more accurate time measurement.
    
    1) t = 7.53-s
    2) t = 8.03-s
    3) t = 7.97-s
    avg. t =7.84-s
    
    t = 7.84 is our measured value.
11. We then measured the percentage of how far off we were from the actual measured value and found that we were off by 0.128% 

CONCLUSION:

We verified the sum-of torques equation, Στ=Iα, by finding the moment of inertia of a certain apparatus and directly applying the equation to predict how long it would take for a cart tied to our apparatus to accelerate down a 1-meter ramp. We compared our prediction with the value we actually measured and we were within 0.128% of the actual value.

A Great Day for Physics.

(11/05/14, 6:58am)