Inertia and torque - Grade: A PDF

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Lab experiment with calculations and data on inertia and torque....

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Torque and Inertia Austin Glass (Lab Partner: Jack McElligott) 3/29/17 ABSTRACT Rotating objects have similar mathematical equations to those of linear bodies. Torque is the product of the moment of inertia and the angular acceleration. This experiment studies torque of dynamic situations to understand moment of inertia better. The experiment was performed by allowing a mass tied to an axle to fall which rotated the axle. Varying masses were attached to the rod threaded through the axle. Error analysis was performed for this experiment to verify theoretical properties. The error ranges were (-0.112kg m2 ≤-0.00723kg m2 ≤ 0.00509kg m2) and the error range for the slopes was (0.000144m2 ≤0.0162m2 ≤0.0199m2). Since the theoretical value did not fall within the error ranges, the experiment failed to accept the theory. Measurement errors most likely caused these results. THEORY Rotating bodies have similar equations to describe their motion as for constant linear acceleration. Where τ is the torque, I is the moment of inertia, and α is the angular acceleration, the equation for the sum of all torques is (eq. 1) Torque does not have to be applied at the radius where the masses are, so its perpendicular distance from the axis of rotation is designated differently from the radius; it is called r. So the torque equation can be written (eq. 2) This concept will be utilized during the experiment. In order to investigate the relationship between the time it takes a mass to fall and the moment of inertia, an equation was derived. The following equation is utilized to show relationship between torque and angular acceleration where m is the mass of the falling weight, t is the time that takes, and d is the distance covered

(eq. 3) Additionally, moments of inertia have the property of being additive, so (eq. 4) where the contribution of the moment of inertia for each single mass element is given by the following equation with M being the total masses on an extended body, and R being the radius the mass is from the center

(eq. 5)

OBJECTIVE The objective of the experiment is to study torque for dynamic situations and to better understand moments of inertia by comparing calculated values. PROCEDURE The experimental apparatus used was similar to that of figure 1. A mass was dropped and allowed to rise up to its maximum height without hitting the floor. This height above the lowest point was measured. The friction was calculated and then g’ was calculated from that. The cross rod was placed onto the rotating apparatus without any masses. A 100 gram mass was released and the following data was recorded: distance, time, mass on cross, radius of central rod, mass of falling weight, radius to center of mass on cross. Those steps were repeated with four different amounts of masses on the cross (both placed 12cm from the center). The mass was adjusted by 50 grams on each side for each different run. The largest mass was run five times in order to use a standard deviation of I for the other parts of the experiment.

DATA Part 1 dstart (m)

T (s)

.959 6.62

dstart (m) T (s) 12.8 0.959 8 16.3 0.959 8

Dmax r (m) m (kg) (m) 0.0063 0 3 0.1 .250

M (kg)

Part 2 M (kg) r (m) m (kg) R (cm) dmax (m) 0.0063 0.122 3 0.100 12.0 0.431 0.0063 0.222 3 0.100 12.0 0.458

0.959 0.959 0.959 0.959 0.959 0.959 0.959

18.6 0 21.5 5 24.8 9 25.4 2 24.9 9 25.4 9 25.4 8

0.0063 3 0.0063 3 0.0063 3 0.0063 3 0.0063 3 0.0063 3 0.0063 3

0.322 0.422 0.522 0.522 0.522 0.522 0.522

0.100

12.0

0.474

0.100

12.0

0.487

0.100

12.0

0.481

0.100

12.0

0.472

0.100

12.0

0.483

0.100

12.0

0.468

0.100

12.0

0.462

CALCULATIONS mg( ds −dm ) ds +dm =0.373N

(0.100kg *9.81m/s/s *(0.959m-0.431m))/(0.959m+0.431m)

Ff =

g’ = g- (Ff/m)

Ik =mr2

9.81m/s/s – (0.373N/0.100kg) =6.25 m/s/s –(N/kg)

t ¿ (¿ ¿ s 2) −1) (g ' 2 ds ¿ ¿¿

0.100kg * (0.00633m)2 *( (

Average I = ´I

(6.25 m/s /s−(N /kg))(12.88 s)2 2 −1) =0.00216kg m 2∗0.959 m

n

=

1 ∑x n i=1 i

Standard deviation of I =σI =

0.00886kg m2= ´I

√

n

∑ ( x´ − x i)2 i=1

0.000114kg m2= σI

n−1

Itr= ML2/12

(0.522kg*(0.342m)2)/(12)=0.00509kg m2

Ia=hπρ(r4/2)

0.233m*π*8000kg/m3*((0.00633m)4/2)=4.70E-6kg m2

Io=Itr+Ia

0.00509kg m2 +4.70E-6kg m2 =0.00509kg m2

Worst/best fit lines=

∆y ∆x

=

y 5 +σ y5 − y +σ y x5 − x1 1

1

(0.00216 6 kg m 2 – 0.0001146 kg m2) – ( 0.008866 kg m2 0.0001146 kg m2) =0.0199 m2 0.122 kg−0.522kg

QUALITATIVE ERROR ANALYSIS A possible cause for error in this lab stemmed from the amount of time it took for the weight to reach its lowest point. Since it was human timed, there was a disconnect between the time that the weight reached the bottom and the time that the timer pressed stop. This may have off put some of the various time data. Another source of error came from wrapping the string up the axle of the apparatus. Precautions were taken to prevent overlap of the string, but it is likely that there may have been some. This would cause a change in the radius, which would have a large impact on the final outcome of the experiment. QUANTITATIVE ERROR ANALYSIS For this experiment values for the moment of inertia were calculated as a mass was released to rotate a rod. For the experiment the average moment of inertia at the M=0.522kg was 0.00886kg m2. The standard deviation was 0.000114kg m2. The worst and best fit lines used the standard deviation of the M=0.522kg for the slope calculations. The calculated slopes for the worst and best fit lines were 0.0199m2 and 0.0162m2.

Ff (N) 0.372 0.347 0.332 0.320 0.324 0.334 0.324 0.338 0.343 Average Ik STDV Ik

RESULTS g' m/s/s –(N/kg) Ik (kg/m ) Itr (kg/m2) Ia (kg/m2) Io(kg/m2) 6.25 0.00216 0.00509 4.70E-06 0.00509 6.51 0.00365 6.66 0.00481 6.78 0.00657 6.72 0.00870 6.64 0.00896 6.74 0.00879 6.60 0.00896 6.55 0.00888 0.00886 0.000114 2

Figure 1. Moment of Inertia Versus the Mass on Rod 0.01 0.01

Moment of Inertia (kg m^2)

0.01 0.01 0.01 0.01 0 0 0 0 0 0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

0.55

Mass on Rod (kg)

Error range: (-0.112kg m2 ≤-0.00723kg m2 ≤ 0.00509kg m2) Slope error: (0.000144m2 ≤0.0162m2 ≤0.0199m2) CONCLUSION The experimental error range for the data was (-0.112kg m2 ≤-0.00723kg m2 ≤ 0.00509kg m2). The error range for the slopes was (0.000144m2 ≤0.0162m2 ≤0.0199m2). The calculated moment of inertia was 0.00509kg m2 which did not fall within the error ranges for the experiment. There was minimal overlap between the best and worst fit lines with the trend line. Since the value was not within the range, the experiment failed to accept the theory. Torque is related to moment of inertia. As the mass began to fall, the rate of rotation of the apparatus increased. This centripetal acceleration meant that a force was applied on the system which may have caused the system to not be in equilibrium. The calculations would not match those experimental values for a system that is not based upon the premises of the mathematical equations. Since the masses were put onto the rod at the top of the apparatus, the radius would be measured from the center of mass of the added masses. The way that the masses were added may have caused a shift in the center of mass which would change the radius for calculations. Also the wrapping of the string about the axle required a measurement of a radius which may have differed from the actual radius during the experiment. Since the calculations depend heavily on the values of radii, slight differences in these values could have significantly altered results....