12 Rotational Equilibrium

Leah Pinkhasov 10-24-2023 University physics 1 Experiment 12 Rotational Equilibrium

Remember that we are using the coordinate system that has the torque bar as the x axis . Your x = 0 point is at the left side of the bar. The y axis is perpendicular to the bar. Calculate torques from the x = 0 point. The sign of the torque is the sign of the force in the y direction (F y ). Forces in the upward direction give positive torques and in the downward direction give negative torques. Part I Fill in the following table. Then find the components and torques of the forces due to the masses to 3 significant digits. (Use g = 10 m/s 2 .) M TorqueBar 0.815 r 1 0.050 r 2 0.050 r 3 0.050 r 4 0.050 Mass (kg) Force=Mg (N) r (m) Angle (°) F x (N) F y (N)  = F y * r (N-m) Torque Bar 0.815 7.99 calculate below -90 0 -7.99 (see below) M1 0.600 5.88 0.412 123 3.20 4.93 2.03 M4 0.450 4.41 0.0635 40 -3.38 2.83 0.180 SUM -0.18 -0.23 2.21 Mass (kg) Force = Mg (N) F x (N) F y (N) τ = F y * r (N-m) Torque Bar 815/1000 0.815 * 9.8 0 (7.99) sin (-90) (see below) M 1 600/1000 0.600 * 9.8 (5.88) cos (123) (5.88) sin (123) (4.93) (0.412) M 4 450/1000 0.450 * 9.8 (-4.41) cos (40) (4.41) sin (40) (2.83) (0.0635) Sum X X 3.20-3.38 4.93+2.83-7.99 2.03+0.180 Calculate the torque of the torque bar and the position of the center of mass. First use equation (4) and solve for  CM (torque of the center of mass). -   = 0 -  1 +  4 +  CM = 0 (This is the sum of all the torques.) Then solve for the position of the center of mass r CM : -  CM = F y oftorque bar * r CM

Part II Fill out a similar table for Part II. The lever arm r of the torque bar = R CM from Part I. Mass (kg) Force (N) r (m) angle (°) F x (N) F y (N)  (N-m) M 4 0.600 5.88 0.0635 62 -2.76 5.19 0.330 M 3 0.200 1.96 0.212 -90 0 -1.96 -0.416 M 2 0.200 1.96 0.152 -90 0 -1.96 -0.298 M 1 0.740 7.25 0.412 110 2.48 6.81 2.81 Torque Bar 0.815 7.99 0.280 -90 o 0 -7.99 -2.24 Sum -0.280 0.0900 0.186 Mass (kg) Force = Mg (N) F x (N) F y (N) τ = F y * r (N-m) M 4 600/1000 0.600 * 9.8 (-5.88) cos (62) (5.88) sin (62) (5.19) (0.0635) M 3 200/1000 0.200 * 9.8 (1.96) cos (90) (1.96) sin (-90) (-1.96) (0.212) M 2 200/1000 0.200 * 9.8 (1.96) cos (90) (1.96) sin (-90) (-1.96) (0.152) M 1 740/1000 0.740 * 9.8 (-7.25) cos (110) (7.25) sin (110) (6.81) (0.412) Torque Bar 815/1000 0.815 * 9.8 (7.99) cos (-90) (7.99) sin (-90) (-7.99) (0.28) sum X X -2.76+2.48 5.19-1.96- 1.96+6.81-7.99 0.330-0.416- 0.298+2.81-2.24 r (m) τ = F y * r (N-m) Equations τ cm = τ F y of torque bar * r CM ∑ τ = 0 τ 1 + τ 2 + τ CM = 0 Torque Bar -2.21 = (-7.99) * r CM r CM = 0.280 m (2.03) + (0.180) + τ CM = 0 τ CM = -2.21

Part III Fill out the table for Part III as you did for Part II . The torque bar is still the x axis. Mass (kg) Force (N) r (m) angle (°) F x (N) F y (N)  (N-m) M 1 0.700 6.86 0.0635 47 4.68 5.02 0.319 M 2 0.200 1.96 0.212 -80 0.340 -1.93 -0.409 M 3 0.200 1.96 0.152 -80 0.340 -1.93 -0.293 M 4 0.700 6.86 0.412 106 -1.89 6.59 2.72 Torque Bar 0.815 7.99 0.280 -80 1.39 -7.87 -2.20 Sum 4.86 -0.120 0.137 Mass (kg) Force = Mg (N) F x (N) F y (N) τ = F y * r (N-m) M 1 700/1000 0.700 * 9.8 (6.86) cos (47) (6.86) sin (47) (5.02) (0.0635) M 2 200/1000 0.200 * 9.8 (1.96) cos (-80) (1.96) sin (-80) (-1.93) (0.212) M 3 200/1000 0.200 * 9.8 (1.96) cos (-80) (1.96) sin (-80) (-1.93) (0.152) M 4 700/1000 0.700 * 9.8 (6.86) cos (106) (6.86) sin (106) (6.59) (0.412) Torque Bar 815/1000 0.815* 9.8 (7.99) cos (-80) (7.99) sin (-80) (-7.87) (0.280) Sum X X 4.68+0.340+0.340- 1.89+1.39 5.02-1.93- 1.93+6.59-7.87 0.319-0.409- 0.293+2.72-2.20 Check: Check ∑f x , ∑f y , ∑ τ in parts 1, 2, 3. All sums should be equal to zero. State clearly how this result conforms to theory. Comment on discrepancies. - The sum of all forces acting on an abject will be zero when at rest due to translational and rotational equilibrium. The sum values of torque should also be zero. During my experiment, my values for part 1, 2, and 3 were all close to zero, but not exactly zero.