W AutoSave Off M25a rotational statics and dynamics 03.16.22 (1) . Saved to this PC v 9 Search (Alt+Q) shemar tulloch ST X File Home Insert Draw Design Layout References Mailings Review View Help Comments Share k Times New Roman ~ 16 ~ A" A Aa Ap EVEEVENT Find Normal Paste BIUvab X x A LA Caption Heading 1 Heading 2 Title Replace Dictate Editor Select v Undo Clipboard Font Paragraph Styles Editing Voice Editor For the cases where the sum of the torques is not zero, the body that has the net torque applied to it is going to experience an angular acceleration. The angular acceleration will be in the same direction as the net torque. This dynamical behavior of rotation has analogous components to the behavior of translational motion where torque ( 7) replaces force, angular acceleration (@) replaces linear acceleration and moment of inertia (1) replaces mass. Again, when the net torque is not equal to zero, the rigid body experiences an angular acceleration in the direction of the net torque. This behavior is described by Newton's second law for rotation: The mass for rotation is replaced by the moment of inertia of the body. It describes not only how much matter a body is composed of, but also how that matter is distributed about the axis of rotation. It is a geometric characteristic of the object, as it depends only on its shape, its mass distribution throughout the shape and the position of the rotation axis. In its discrete form it is defined by: 1 = Emr This expression means take each particle of mass multiplied by its radius squared (where the radius is the distance from the center of the mass to the center of the axis of rotation) and add them all up to get the total. I = miri + marz + mar's + mark .mark For a disk with radius (R) and mass (M) rotating about (z) axis, the moment of inertia can be calculated by: ! MR' Figure 5 For a thick-walled cylindrical tube with open ends, inner radius (71), outer radius (12) and mass (m) rotating about (z) axis, the moment of inertia can be calculated by: OneDrive . . . X Screenshot saved The screenshot was added to your OneDrive. Date Modified 03/16/22 Page 4 of 13 2640 words X Text Predictions: On x Accessibility: Investigate " Focus 70%W AutoSave Off M25a rotational statics and dynamics 03.16.22 (1) . Saved to this PC v 9 Search (Alt+Q) shemar tulloch ST X File Home Insert Draw Design Layout References Mailings Review View Help Comments Share k Times New Roman ~ 16 ~ A" A Aa Ap Find Heading 2 Normal Paste Caption Heading 1 Title Replace BIUvab X x A LA Dictate Editor Select v Undo Clipboard Font Paragraph Styles Editing Voice Editor I, = =m(r +r ) Figure 6 The angular acceleration of the rigid body can be related to the linear acceleration (may be called tangential acceleration) at some point, a radial distance out from the axis of rotation. For cases like this experiment where the system is rotating with a constant angular acceleration (of), it is possible to calculate the linear acceleration (@) for a point located at distance (R) from the center of rotation using the following formula a = Ra The rotation in this experiment is produced by an external applied torque. The force for this torque is from the tension in an attached string (Fr) due to a mass hanging on the other end. The weight of the mass is greater than the tension in the string and results in the mass moving with acceleration (a) downward. The magnitude of the tension can be calculated by the following F, = m(g - a) T AFT Fnet = FT - W = -ma FT = W -ma VW = mg OneDrive . . . X Figure 7 Screenshot saved The screenshot was added to your OneDrive. Date Modified 03/16/22 5 Page 5 of 13 2640 words X Text Predictions: On x Accessibility: Investigate " Focus 70%W AutoSave Off M25a rotational statics and dynamics 03.16.22 (1) . Saved to this PC v 9 Search (Alt+Q) shemar tulloch ST X File Home Insert Draw Design Layout References Mailings Review View Help Comments Share k Times New Roman ~ 16 ~ A" A Aa Ap Find Paste Caption Heading 1 Heading 2 Normal Title Replace BIUvab X X A LA Dictate Editor Select v Undo Clipboard Font Paragraph Styles Editing Voice Editor Procedures: This experiment consists of two separate parts. The first part is conducted in order to calculate torque and verify rotational equilibrium using the meter stick and the set of masses as shown in Figure 2. The second part is conducted in order to examine rotational dynamics by varying the torque applied to the disk and hoop, determining the resulting acceleration and ultimately calculating the moment of inertia. This part uses the apparatus shown in Figure 3. Part I: Rotational Equilibrium 1. Record the position of the point of suspension where the meter stick is balanced with no masses attached to it. This represents the center of mass of the meter stick, and can be found by looking at the center, point of suspension, of the clamp. The position should be recorded using units of meters with precision to the millimeter in Table 1. 2. For each trial measure the total mass placed on each side (combined mass and mass hanger) using the most precise balance available. Remember to convert to kilograms. 3. Place a 100-gram mass and hanger on one side of the meter stick by hanging it from the string at some convenient location. Hang a 50-gram mass and hanger on the other side of the meter stick, again from the string but now moving it along the stick until the meter stick reaches equilibrium and is level to the table. Record the position for each mass, using meters, in Table 1. 4. Calculate the lever arm for each mass. This is the distance from the mass to the point of suspension. Next calculate the force and then the torque for each side. Note: the system has reached equilibrium; therefore both values for torque should be identical, or at least very close. 5. For trial 2 place one of the known masses, from the mass kit, on the left side and the unknown mass provided for the right side. Note: Make sure the masses differ from those used in trial 1 Again position them so equilibrium is reached. 6. Complete all the calculations for your known mass, entering the information on Table 2. 7. For the unknown mass, record its position and lever arm. Since we are dealing with a mass of unknown quantity, the torque cannot be directly calculated. But since the system is at equilibrium, we can assume their torques to be identical. Record the same torque value obtained for the known mass as also being the unknown's torque 8. Use this torque to calculate the force and then the mass of the unknown. Record your results on Table 2. Date Modified 03/16/22 Page 7 of 13 2640 words X Text Predictions: On x Accessibility: Investigate " Focus + 70%W AutoSave Off M25a rotational statics and dynamics 03.16.22 (1) . Saved to this PC v 9 Search (Alt+Q) shemar tulloch ST X File Home Insert Draw Design Layout References Mailings Review View Help Comments Share k Times New Roman ~ 16 ~ A" A Aa Ap Find Caption Heading 1 Heading 2 Normal Paste Title Replace BIUvab X X A LA Dictate Editor Select v Undo Clipboard Font Paragraph Styles Editing Voice Editor Part II: Rotational Dynamics Varying the Radius - Disk 1. The step pulley under the disk has three different radii. Measure each diameter and calculate the radius converting them into units of meters. Record the data in Table 3. 2. Mount the large disk above the step pulley on the center shaft aligning the flat edge of the shaft with the disk. 3. Add 100-grams to a mass hanger and measure the total mass of the combination. The mass will remain constant during this part. Convert the mass to kilograms and record the data in Table 3 4. Attach the mass hanger to the string. Wind the string on the largest radius until the mass hanger is suspended close to the outside pulley. 5. Adjust the position of external pulley so that the string is aligned with the rod of the pulley. 6. If necessary ask a Lab Assistant to provide computer related instructions to begin the computer data collection and analyses. The computer will display a graph of angular velocity over time. Obtain the statistical slope of the line to find the angular acceleration and its standard deviation for each trial. Make sure to collect at least five significant digits for the angular acceleration. Note: when conducting each trial, press start to begin collecting data first and then release the mass. Press stop when the mass hanger reaches the floor Repeat these steps using the other two radii. Record all of your data in Table 3. Varying the Force - Disk 8. Place a 75-gram mass on the mass hanger and measure the total mass. The radius will now remain constant during this part of the experiment using only the middle radius. Wind the string until the mass hanger is suspended close to the outside pulley adjusting its position so it is aligned with the mounting rod, as before. 10. Obtain the angular acceleration and its standard deviation as in the above trials. Record the data in Table 3. 11. Conduct one more trials, using 125 grams with the mass hanger. Again obtain the angular acceleration and standard deviation as before. Record your data in Table 3 Varying the Radius and the Force - Disk and Hoop 12. Next add the hoop on top of the disk being carful center it in the middle of the disk. 13. Repeat the sequence of steps from 2 through 11 as above but now with the hoop added to the top of the disk for each trial. Record all of this data in Table 4. OneDrive . . . X Theoretical Moments of Inertia 1. Measure the diameter of the disk. Screenshot saved 2. Measure the inside and outside diameter of the hoop. 3. Measure the mass of the disk and the hoop separately. The screenshot was added to your OneDrive. Date Modified 03/16/22 8 Page 8 of 13 2640 words X Text Predictions: On x Accessibility: Investigate " Focus 70%W AutoSave Off M25a rotational statics and dynamics 03.16.22 (1) . Saved to this PC v 9 Search (Alt+Q) shemar tulloch ST X File Home Insert Draw Design Layout References Mailings Review View Help Comments Share k Times New Roman ~ 16 ~ A" A Aa Ap Find Paste Caption Heading 1 Heading 2 Normal Title Replace BIUvab X X A LA Dictate Editor Select v Undo Clipboard Font Paragraph Styles Editing Voice Editor Analysis: Part II: Rotational Dynamics 1. Calculate the linear acceleration (@), tension (Fr) and torque (T) for the first trial in Table 3. After you have the values calculated for the first trial have a lab instructor confirm the result. 2. The lab instructor will guide you on the remaining calculations for Tables 3 & 4. For the first graph of torque vs angular acceleration use the data on Table 3. Put the torque data on the y-axis and the angular acceleration data on the x-axis. This data should represent a linear elationship. The slope of the statistical linear regression line will be the experimental moment of inertia for the disk. For the second graph of torque vs angular acceleration use the data on Table 4. Put the torque data on the y-axis and the angular acceleration data on the x-axis. Again this data should represent a linear relationship. The slope of the statistical linear regression line will be the experimental moment of inertia for the disk and hoop. 5. Calculate the experimental moment of inertia for the hoop by taking the difference between the two moments, disk and hoop minus disk. 6. Using the dimensions and masses measured for the disk and the hoop; calculate the theoretical moments of inertia for each. Also calculate the percent error for each comparing the experimental value to the theoretical value. OneDrive ... X Screenshot saved The screenshot was added to your OneDrive. Date Modified 03/16/22 Page 9 of 13 2640 words X Text Predictions: On x Accessibility: Investigate " Focus 70%W AutoSave Off M25a rotational statics and dynamics 03.16.22 (1) . Saved to this PC v 9 Search (Alt+Q) shemar tulloch ST X File Home Insert Draw Design Layout References Mailings Review View Help Comments Share k Times New Roman ~ 16 ~ A" A Aa Ap Find Paste BIUvab x x ALA= = ~ ~ Caption Heading 1 Heading 2 Normal Title Replace Dictate Editor Select v Undo Clipboard Font Paragraph Styles Editing Voice Editor Data Sheets: M25a: Rotational Equilibrium and Rotational Dynamics NAME: DATE: Table 1: Rotational Equilibrium Location of point of suspension = 0.498 (m) Trial One mass (kg) position (m) lever arm (m) force (N) torque (Nm) left side 0.054912 0.211 right side 0.104642 0.399 Table 2: Rotational Equilibrium - Unknown Mass Unknown mass (recorded from lab balance) = 0.041263 (kg) Trial Two mass (kg) position (m) lever arm (m) force (N) torque (Nm) left side 0.034956 ).258 position (m) lever arm (m) torque (Nm) force (N) calc. mass (kg) right 0.699 side % Difference between calculated and measured mass = Date Modified 03/16/22 11 Page 11 of 13 2640 words X Text Predictions: On Accessibility: Investigate " Focus + 70%W AutoSave Off M25a rotational statics and dynamics 03.16.22 (1) . Saved to this PC v 9 Search (Alt+Q) shemar tulloch ST X File Home Insert Draw Design Layout References Mailings Review View Help Comments Share k Times New Roman ~ 16 ~ A" A Aa Ap Find Paste BIUvab X X A LA Caption Heading 1 Heading 2 Normal Title Replace Dictate Editor Select v Undo Clipboard Font Paragraph Styles Editing Voice Editor Data Sheets: M25a: Rotational Equilibrium and Rotational Dynamics NAME: DATE: Table 3: Rotational Dynamics - Disk applied mass radius angular stand. acc. dev. linear acc. tension torque 0.104882 0.00829 0.79303 .005 varying radius 0.104882 0.012485 1.2605 .005 0.104882 0.018695 1.907 .005 applied radius angular stand mass acc dev. linear acc. tension torque 0.07991 varying mass 0.012485 0.9443 .005 0.129891 0.012485 1.567 005 Table 4: Rotational Dynamics - Disk and Hoop applied angular stand mass radius acc dev. linear acc. tension torque 0.104882 0.00829 0.53763 .005 varying radius 0.104882 0.012485 0.80127 .005 0.10488 0.018695 1.2423 005 applied stand. mass radius angular ace dey linear acc tension torque OneDrive ... X 0.07991 varying mass 0.012485 0.59773 005 Screenshot saved 0.129891 0.012485 1.024 .005 The screenshot was added to your OneDrive. Date Modified 03/16/22 12 Page 12 of 13 2640 words DX Text Predictions: On Accessibility: Investigate " Focus 70%W AutoSave Off M25a rotational statics and dynamics 03.16.22 (1) . Saved to this PC v 9 Search (Alt+Q) shemar tulloch ST X File Home Insert Draw Design Layout References Mailings Review View Help Comments Share k Times New Roman ~ 16 ~ A" A Aa Ap Find Normal Paste Caption Heading 1 Heading 2 Title Replace BIUvab X X A LA Dictate Editor Select v Undo Clipboard Font Paragraph Styles Editing Voice Editor Data Sheets: M25a: Rotational Equilibrium and Rotational Dynamics NAME: DATE: Experimental Determination Moments of Inertia: Moment of Inertia for the Disk: Moment of Inertia for Disk and Hoop: Moment of Inertia for the Hoop: Theoretical Determination: Diameter of the Disk: 0.228 Mass of the Disk: 1.4303 Theoretical Moment of Inertia of the Disk: Percent Error: % Inside Diameter of the Hoop: 0.10669 Outside Diameter of the Hoop 0.12629 Mass of the Hoop: 1.473 Theoretical Moment of Inertia of the Hoop: Percent Error: Date Modified 03/16/22 13 Page 13 of 13 2640 words X Text Predictions: On Accessibility: Investigate " Focus + 70%W AutoSave Off M25a rotational statics and dynamics 03.16.22 (1) . Saved to this PC v 9 Search (Alt+Q) shemar tulloch ST X File Home Insert Draw Design Layout References Mailings Review View Help Comments Share k Times New Roman ~ 16 ~ A" A Aa Ap Find Paste Caption Heading 1 Heading 2 Normal Title Replace BIUvab X X A LA Dictate Editor Select Undo Clipboard Font Paragraph Styles Editing Voice Editor M25a: Rotational Equilibrium and Rotational Dynamics O. Introduction: By Newton's First Law of Motion, a body is in translational equilibrium when the net force on the object is equal to zero. However, a net force of zero can still allow an object to rotate. The tendency of a force to rotate a body about an axis is called Torque. Therefore when examining a body that could rotate and again applying Newton's First Law, the body is in rotational equilibrium when the net torque is equal to zero. For translational equilibrium the body will be either not moving or moving in a straight line at constant velocity. Similarly for rotational equilibrium the body is either not rotating or is rotating with constant angular velocity. If the body is completely stationary, meaning both not moving along a line and not rotating, then it is in static equilibrium. Newton's Second Law of Motion explains what happens when the sum of the Forces and/or the sum of Torques are not equal to zero. In these cases the body experiences acceleration as a result of the net force and/or an angular acceleration as a result of the net torque. The Second Law, in rotational form, establishes the relationship between the net torque, the moment of inertia and the angular acceleration. Moment of Inertia represents the mass for rotation. It refers not only to how much mass an object has but also to how this mass is distributed with respect to the axis of rotation. The main purpose of this experiment is to examine the properties of torque, the dynamics of rotation that occur as a result of a net torque and the moment of inertia for a rigid body. First, multiple torques will be applied to a meter stick, putting it in static equilibrium. Then, the individual torques will be determined and used to verify the first law condition for equilibrium. For the dynamic part of the experiment torque will be applied to an apparatus composed primarily of a large horizontal solid disk and a heavy thick hoop that will mount on top of the disk. The disk or disk & hoop combination will accelerate angularly due to the net torque. The data collected will be used to determine the moment of inertia for both bodies. This experimentally determined moment of inertia will be compared to the theoretical value expected. Apparatus: Figure 1 Date Modified 03/16/22 Page 1 of 13 2640 words X Text Predictions: On x Accessibility: Investigate " Focus + 70%W AutoSave Off M25a rotational statics and dynamics 03.16.22 (1) . Saved to this PC v 9 Search (Alt+Q) shemar tulloch ST X File Home Insert Draw Design Layout References Mailings Review View Help Comments Share k Times New Roman ~ 16 ~ A" A Aa Ap Find Caption Heading 1 Heading 2 Normal Paste Title Replace BI Uvab X x A LA Dictate Editor Select v Undo Clipboard Font Paragraph Styles Editing Voice Editor Part One: Rufe in equilibrium Meter stick Support stand Clamp to hold the meter stick Set of masses 2 Mass hangers ... .. Unknown mass Hangers Assorted masses Figure 2 Part Two: Height-adjustable stand with center mounted bearing shaft 3-step pulley Large solid disk Heavy thick hoop >Pulley with mounting rods Mass hanger, masses and string let of Masses >Caliper and metric rule Photogate with computer timing system Figure 3 Page 2 of 13 2640 words X Text Predictions: On x Accessibility: Investigate " Focus + 70%W AutoSave Off M25a rotational statics and dynamics 03.16.22 (1) . Saved to this PC v 9 Search (Alt+Q) shemar tulloch ST X File Home Insert Draw Design Layout References Mailings Review View Help Comments Share k Times New Roman ~ 16 ~ A" A Aa Ap Find Normal Paste Title Replace BIUvab X X A LA Caption Heading 1 Heading 2 Dictate Editor Select v Undo Clipboard Font Paragraph Styles Editing Voice Editor Discussion: Torque is the tendency of a force to rotate a body about an axis. It depends on the magnitude of the force, the point where the force is applied relative to the axis of rotation, and the direction of the force. It is defined by the vector equation: |=\\R|F|sin & T = Torque R = radius F = force a = angle (between force and radius) This expression can sometimes be simplified by using the effective lever arm. The lever arm represents the distance between the force's line of action and the axis of rotation (suspension point). It is measured on a line that is perpendicular to both (Figure 4). The lever arm's equivalency is given by the following expression: L = R|sin e Suspension point Position Position Lever Arm L Lever ArmR Mass MassR F. F . Figure 4 Combining these two expressions, the magnitude of the torque can be calculated using: 171 = |FIL Where the magnitude of the applied force on this situation (Figure 4) is the force of gravity of the hanging mass: IF| = mg OneDrive . . . X When examining the direction of the torque, it is usually considered positive when the force tends to produce a counterclockwise rotation about the axis, and negative when the force tends to produce a clockwise rotation. If a rigid body is acted upon by a system of torques, where the sum of these Screenshot saved torques is zero, the rigid body is in equilibrium with respect to rotation. This means that the body The screenshot was added to your can only have two rotational motion states: to be at rest or to rotate uniformly (constant angular speed) about a fixed axis. OneDrive. Date Modified 03/16/22 Page 3 of 13 2640 words X Text Predictions: On x Accessibility: Investigate " Focus 70%