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Moment of Inertia Calculator
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Item to guess, measure, or calculateEnter formulas hereCorrect AnswersHelpful Notes:
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Guess what goes in this blank... The moment of inertia of this object can be calculated using the formula
I = ____ mr^2.		NANAFor a thin hoop, I = mr^2. For a disc, I = 0.5 mr^2. For a sphere, I = 0.4 mr^2. A thin hoop has the highest moment of inertia, because all of its mass is at its edge. A sphere has the lowest, because a lot of its mass is close to its center. The goal here is to guess how this wheel and axle's mass distribution compares to the other objects. Based on the mass distribution of this object, what multiple of mr^2 would you guess is equal to its moment of inertia? You will eventually be calculating this multiplier in cell B22.
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Wheel and Axle Mass (kg)0.350.35Measure with the balance. Don't forget to convert to kg.
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Wheel radius -- radius at outermost point (m)0.060.06Measure with a ruler. Conver to meters!
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Axle Radius (m)0.0050.005These axles are supposed to be 0.25 inches. Google the conversion to m.
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Mass of falling weight (kg)0.10.1Confirm by placing it on the balance. Don't forget to convert to kg.
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Wrapped String length (m)0.70.7This is also the distance that the weight drops before falling free. Measure only the wrapped string. Use a meter stick.
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Acceleration time (s)55Repeat this step until you do it well. Use a timer with a split function, and practice with your split function ahead of time. Start the timer when the weight begins to fall. End the first split when the thread comes free. Stop the second split when the axle stops moving. Record the first split here.
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Deceleration time (s)2525Record the 2nd split (time for axle to come to rest from its max speed).
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Falling Weight acceleration during string pull -- absolute value (m/s/s)0.0560.056Use good ol' linear kinematics. You know the weight's initial velocity. You know the distance it accelerated while pulling the string. You know the duration of this event (first split). Find the acceleration. As soon as you enter your formula here, look at columns E and F to see if your formula works like mine. I have copied your formula from column B into column E. If your formula is correct, your column E answer should match my column F answer.
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Wheel and axle angular acceleration during string pull -- absolute value (rad/s/s)11.211.2Convert the linear acceleration that you just calculated to angular acceleration. There's an easy formula.
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String tension during string pull (N)0.97440.9744Remember back to the forces unit. The mass is hanging from the string, experiencing two forces; weight and tension. Net force = ma = the sum of the forces acting on the mass. Solve for tension. String tension should be a little less than the weight of the falling mass, since it's accelerating downward.
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Torque applied by string during fall -- absolute value (Nm)0.0048720.004872Basic torque formula. The first one you learned in this unit.
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Max angular speed of wheel and axle (rad/s)5656This is the angular speed of the wheel and axle at the end of the string pull interval. Use kinematics, but the new rotational kind. You know duration of the string pull interval. You also know the angular acceleration and the initial angular velocity for this interval.
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Angular acceleration of wheel and axle during slowing period -- absolute value (rad/s/s)2.242.24This interval begins when when the string pull ends (the string comes loose). So you know the starting angular velocity for this interval. You know the elapsed time. You also know the final angular velocity for the slowing down period.
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Wheel and Axle Moment of Inertia (kgm^2)0.00036250.0003625To find these two quantities, you'll need to set up a simple system of 2 equations. For both equations, use the formula that is the "rotational version of F=ma." For one equation, apply this formula to the string pull period -- when the wheel and axle is speeding up. During this period, the net torque equals the torque from the string tension minus the torque from friction (because both contribute to the net torque). Make sure that you use the correct angular acceleration for the string pull interval. You have already calculated the torque applied by the string and the angular acceleration. Your two unknown variables will be the frictional torque and the moment of inertia.

For the other equation, apply the same formula to the slowing down period -- after the string releases. For the slowing down period, the net torque is equal to only the torque from friction (because the string is no longer pulling). In this equation, remember to use the correct anngular acceleration for the slowing down period. Again, your two unknown variables will be the frictional torque and the moment of inertia.

Now you have two equations and two unknowns. Find the moment of inertia and the frictional torque.
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Torque due to friction -- absolute value (Nm)0.0008120.000812
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Wheel and Axle mr^2 (kgm^2)0.001260.00126This is preparation for the next box. Just do what it says. Multiply the mass of the wheel and axle (in kg) by the square of its radius where its radius is the greatest. Make sure that you're using meters for the axle radius units.
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For this particular mass distribution, I = ______mr^2 (no units)0.28769841270.2876984127This is a fill-in-the-blank question. I = ____ mr^2. You've already calculated I and mr^2, so now this is a very easy algebra problem. When you're done, compare your calculation with your earlier guess (which you entered in cell B4).
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