Moment of Inertia

Moment of Inertia
In general, when a force is applied to a body, the motion of the system can be
broken down into translational motion of the center of mass and rotational
motion about the center of mass. In this experiment, a weight is attached to a
cylinder that will rotate about its axis as the weight descends (Fig. 1). Since
the axis is fixed, there is no translational motion.
Newton’s second law will also apply to rotating body. However, the law is
simplified if the variables are rotational variables. Newton’s second law
becomes
τ = Iα
(1)
where τ is the torque expressed in dyne-cm or N-m, I is the moment of inertia
in g cm2 or Kg m2 , and α is the angular acceleration in radians/sec2.
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The moment of inertia, like mass in linear motion, is the inertial property of the
system which opposes a change in angular motion. It is possible to calculate
the moment of inertia of a body if the distribution of the mass is known. The
moment of inertia of objects used in this experiment is given as follows:
R
R
Solid Cylinder I = ½ MR2 (2)
Thin Ring I = MR2 (3)
For the experimental set up as shown in Figure 1, the linear acceleration (a) of a
descending weight (mg) is measured by using a smart pulley. If the string does not slip
on the step pulley, then the linear acceleration of the weight will be the same as the
tangential acceleration of a point on the rim of the step pulley. The tangential
acceleration is related to the angular acceleration by Equation 5.
(5)
α=
a
r
(5)
*where r is the radius of the step pulley the
string is wrapped around.
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1
As shown in Fig. 1, the tension (T) of Fig.
the string
produces a torque (T r) on the
cylinder. The equation of motion of the cylinder is given by Eqn. 6. Likewise
Eqn 7 is the equation of motion for the descending mass.
(6)
(7)
τ = Tr = Iα
mg − T = ma
Solving Eqn 7 for T and substituting into Eqn. 6 will give an equation for the
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torque (τ).
τ = mr(g − a)
(8)
If torque (Eqn. 8) is plotted versus angular acceleration (Eqn. 5), then a straight
line whose slope is the moment of inertia should result. The radius r in the
equations is the radius of the pulley the string is wrapped around, not the radius of
€the larger platter.
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Procedure
1.
Record the mass of the hanger and the radius of the step pulley. Thread
the string through the holes in the step pulley so that the pulley will wind
up on the middle pulley. Stretch the string over the smart pulley and
attach it to a hanging mass. Make sure the string is as parallel to the disc
as possible. Open Logger Pro/ Open/ Experiments/Probe and
Sensors/Photogates/ Pulley.xmlb.
2.
Place 40 g on the 5g weight hanger. On the computer screen, click the
mouse on the "START" button and release the cart. Click "STOP" on the
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computer screen when the weight hits the floor.
3. Click on the "Velocity Graph", then click on the R= icon at the top of the
graphs. The computer program automatically calculates the slope and
draws the best fitting straight line through the data points. The slope of
the graph is the acceleration of the mass. Record this value in the data
table.
4.
Repeat the procedure in steps (2) and (3) with 80g, 120g, 160g and 200g
on the 5g-weight hanger.
5.
Measure the mass of cylinder and its radius. Calculate its moment of
inertia using Equation 2. Show your work in the data section. Place the
cylinder on top of the spindle and repeat the five measurements done in
steps 2 - 5 above.
6.
Measure the mass of ring and its outside radius and inside radius. The
average radius is the sum of the outside radius and inside radius divided
by two. Calculate its moment of inertia using Eqn. 3. Show your work
in the calculation section. Take the cylinder off and place the ring on top
of the spindle. Repeat the five measurements done in steps 2 - 5 above.
7.
Calculate the torque using Equation 8 and the angular acceleration using
Equation 5. Record these values in the tables.
8.
Plot torque versus angular acceleration for disc with step pulley only, the
disc with step pulley plus the cylinder, and the disc with step pulley plus
the ring. In each graph, use a linear curve fit to find the slope of the curve
and click “set the intercept to zero box” (The box is just above the
“display equation on graph box”. The slope of the curve is the moment of
inertia of the system. Therefore, the experimental value of the moment of
inertia of the cylinder is just the difference between the slope for the disc
with the step pulley plus the cylinder and the disc with step pulley only.
Find the experimental values for the moment inertia of the cylinder disk
and the ring. Find a percent error between the experimental value and the
calculated value.
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Data:
1. Disc with the step pulley
Radius of step pulley (Middle pulley) = ______________________
Hanging Mass
Acceleration Angular Acceleration (α)
Torque (τ)
1.
2.
3.
4.
5.
Moment of inertia of Step Pulley (Slope of graph) = ____________________
2. Disc with step pulley plus solid cylinder
Mass of Cylinder = _______________
Radius of Cylinder (R)= _____________
Calculated Moment of Inertia = _______________ Slope = ___________________
Exp. value of Moment of Inertia of the cylinder = ________________
% error = ____________
Hanging Mass
Acceleration Angular Acceleration (α)
1.
2.
3.
4.
5.
Show work for the calculated moment of inertia of a cylinder
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Torque (τ)
3. Disc with step pulley plus ring
Mass of Ring = ________ Outside Radius= ________
Inside Radius = __________
Average Radius of Ring (R) = _________________
Calculated Moment of Inertia = ___________________
Slope = _______________
Exp. value of Moment of Inertia of the ring = __________
Hanging Mass
Acceleration
% error = _____________
Angular Acceleration (α)
1.
2.
3.
4.
5.
Show work for the calculated moment of inertia of a ring
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Torque (τ)