P07_Rotational Motion

Old Dominion University PHYS 111 & PHYS 226/231/261 Lab Manual 1 OLD DOMINION UNIVERSITY PHYS 111 PHYS 226/231/261 P07 – ROTATIONAL MOTION Submitted By: 1. 2. 3. Submitted on Date Lab Instructor

2 Experiment P07: Rotational Motion Rotational Motion Experiment P07 Objective  Investigate the conditions needed for a rigid body to remain in static equilibrium  Investigate the effect of distribution of mass on rotation Materials Universal Table Clamp Paper Clips, Small (1 box) Right Angle Clamp Equal Arm Balance Mass Hanger, Small, 50 g (2) Torque Chain (2) Unknown Slotted Mass (Colored Orange) Lab Support Rod, 36” Slotted Mass Set Digital Scale Meter Stick Vernier Calipers Rotational Inertia Apparatus Stopwatch Thread Scissors Theory Torque: In a previous experiment, "Resolution of Forces," you investigated a system of three forces whose lines of action were concurrent (intersected in a single point.) This illustrated the First Law of Static Equilibrium , which states that for a body to remain stationary, the sum of the forces acting on the body must be zero . It was not necessary to consider the torque (or turning effect) in this system of concurrent forces. This experiment is designed to show clearly how torques—turning forces—are produced when the forces acting on a body are not concurrent . Forces will be applied to an Equal Arm Balance which is free to pivot, and which will turn when an unbalanced torque is produced. The situation will be simplified by keeping all forces parallel so that we can concentrate on the behavior of torques without being concerned at the same time with the vector addition of forces. The equation for calculating torque is: τ = r ×F (1) where is the torque, τ r is the lever arm and F is the applied force. The Second Law of Static Equilibrium : The sum of all the torques produced by the various forces acting on a body around any one chosen pivot point must be zero for static equilibrium to occur . A torque is defined as positive if it tends to produce a counterclockwise rotation around the pivot point and negative for a clockwise rotation. Together, the two Laws of Static Equilibrium represents the conditions that cause a rigid body to neither move nor rotate. They provide the basis for the design of structures that must withstand stresses and loads. Moment of Inertia: The resistance to any change in rotation of a body is called its rotational inertia or moment of

4 Experiment P07: Rotational Motion Left Loop Right Loop Setup 1 L1 R1 with 100 g Setup 2 L2 R1 with 100 g Setup 3 L3 with 100 g R1 Setup 4 L3 with 100 g R2 Setup 5 L2 with Unknown Mass (orange mass) R3 Note: When the table says “with 100 g” that means to add 100 grams to that hanger. (In your data tables, be sure to account for the mass of the hangers/chains in your totals.) You are then to determine how much mass it takes to add to the other hanger to achieve balance. Part B: Energy Transfer The goal of this experiment is to calculate the moment of inertia of a system by a dynamical method and compare it with an estimate based on point mass approximations. For studying rotational inertia, a system will be used that consists of two masses located at opposite ends of a crossbar attached at the center to a vertical shaft which is mounted in a housing that allows it to rotate freely (see Figure P07.3). A string is wrapped around the shaft and drawn over a pulley to a hanging mass, M. As this mass drops through a known height h , it applies a torque to the main shaft and the mass's gravitational potential energy is transformed into the kinetic energy of the system.

Old Dominion University PHYS 111 & PHYS 226/231/261 Lab Manual 5 The moment of inertia for the system (consisting of the shaft, crossbar, masses and wing nuts) can be found from conservation of energy. The potential energy change of the falling mass is the sum of the kinetic energy changes of the falling mass and the rotational system. In the following equation M is the falling mass and 'rot' is the rotating portion of the system: PE M = KE M + KE rot (3) Using parameters that can be measured this becomes: Mgh = 1 2 M v 2 + 1 2 I ω 2 (4) Again, M is the falling mass, g is the acceleration due to gravity, h is the height the mass falls through, v is the final velocity of the falling mass, I is the moment of inertia of the entire rotating system, and ω is the final angular velocity of the rotating system. The angular velocity ( ) of the system can be related to the linear velocity ( ω v ) of the falling mass by noting that the distance traveled by the falling mass is the same as the distance traveled by a point on the circumference of the shaft of radius r . Hence ω = v / r , where r is the radius of the rotating shaft that has the string wrapped around it. Further, mass falling under constant acceleration will have a final velocity of v final = 2 v average = 2 h / t . Now (and this is left for students to show the math steps) equation (4) becomes: I = M r 2 ( gt 2 2 h − 1 ) (5) Equation (5) gives the moment of inertia of the entire system that is rotating. To determine the moment of inertia of the two masses alone ( I m ), the experiment is done both with and without the masses (I t and I 0 respectively in equation (6)). This is similar to determining the mass of water in a beaker by measuring the beaker's mass with and without water and then just subtracting out the mass of the beaker. I m = I t − I 0 (6) I m = [ M r 2 ( g t 2 2 h − 1 ) ] t − [ M r 2 ( gt 2 2 h − 1 ) ] 0 (7) I m = M r 2 g 2 h ( t t 2 − t 0 2 ) (8) The last equation provides an experimental method of determining the moment of inertia the two masses spinning on the crossbar by knowing the mass hanging on the string M , the height the mass

Old Dominion University PHYS 111 & PHYS 226/231/261 Lab Manual 7 Data Table – Experiment P07 Data Table 1 Equal Arm Balance Pre-Measurements Lever Arm Loop L3 Loop L2 Loop L1 Loop R1 Loop R2 Loop R3 m m m m m m Hanger Assembly Mass Left Assembly Chains/Hanger kg Right Assembly Chains/Hanger kg Data Table 2 Loops L1 & R1 Loop Lever Arm ( r ) Mass Force Torque ( τ ) L1 m kg N N  m R1 m kg N N  m Net Torque N  m Data Table 3 Loops L2 & R1 Loop Lever Arm ( r ) Mass Force Torque ( τ ) L2 m kg N N  m R1 m kg N N  m Net Torque N  m Data Table 4 Loops L3 & R1 Loop Lever Arm ( r ) Mass Force Torque ( τ ) L3 m kg N N  m R1 m kg N N  m Net Torque N  m Data Table 5

8 Experiment P07: Rotational Motion Loops L3 & R2 Loop Lever Arm ( r ) Mass Force Torque ( τ ) L3 m kg N N  m R2 m kg N N  m Net Torque N  m Data Table 6 Determining an Unknown Mass – Loops L2 & R3 Loop Lever Arm ( r ) Mass (kg) Force (N) Torque ( τ ) L2 m R3 m N  m Unknown Mass Summation Mass of Unknown Mass, direct measurement kg Mass of Unknown Mass, experimental measurement kg Percent Difference % Data Table 7 Energy Transfer Trials with Masses on Crossbar (I t ) Trials without Masses on Crossbar (I 0 ) Trial Time t t Average t t Trial Time t 0 Average t 0 1 s s 1 s s 2 s 2 s 3 s 3 s 4 s 4 s 5 s 5 s Data Table 8