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Procedure: To find center of mass of the ruler Experimental setup: Part A: Prove Sum of Torques - 0 Step 1 Step 2 Step 3 Step 3a Step 1: Gently insert one end of ball bearing at the center of the Styrofoam cup as shown in step 1 Step 2: Apply tape at the center of the ruler followed by placing ruler on the ball bearing, such that the ruler is balanced. Step 3, 3a: Record center of mass, the point where the ruler is balanced is the center of mass of ruler, record the observation (indicated by arrow) The center of mass of the ruler - re . cm= m CHNLES LEON je, o Step 4 Step 4: Hang m, = 50g and m,-100g mass on the ruler with a string (see fig above) followed by gently moving m, or m, left or right on ruler, such that the meter stick is balanced as in step 2 Step 5: Find the distance r, and re from re, record data in observation table below. Example: In figure below, r2=- 20 + 50 - 30 cm, r, - 87 - 50 - 37 cm 60 100 F=mag Styrofoam cup F:-mig Find force Find force F2of the F1 of the hanging hanging mass mass Fig 1: Experimental set up Step 6: Find Force on each end of the ruler by substituting vales for m, and m2 in equation F-mag and F=mag Step 7: Exchange weights m, and mą and repeat from step 4 Step 8: There is no step 8 Distance from Torque center of ruleri= Fr (1cm - 0.01m) Sum of F- mg (N) g- 9.8 m/s² Mass (m) in Kg Obs # Torques (1g-0.001kg) i= 11+ 12 0.05 r = m 1 F2 m2 0.10 : =-m F 0.10 r1 = m m2 0.05 12 = - m Is sum of torque zero?

Procedure: To find center of mass of the ruler Experimental setup: Part A: Prove Sum of Torques - 0 Step 1 Step 2 Step 3 Step 3a Step 1: Gently insert one end of ball bearing at the center of the Styrofoam cup as shown in step 1 Step 2: Apply tape at the center of the ruler followed by placing ruler on the ball bearing, such that the ruler is balanced. Step 3, 3a: Record center of mass, the point where the ruler is balanced is the center of mass of ruler, record the observation (indicated by arrow) The center of mass of the ruler - re . cm= m CHNLES LEON je, o Step 4 Step 4: Hang m, = 50g and m,-100g mass on the ruler with a string (see fig above) followed by gently moving m, or m, left or right on ruler, such that the meter stick is balanced as in step 2 Step 5: Find the distance r, and re from re, record data in observation table below. Example: In figure below, r2=- 20 + 50 - 30 cm, r, - 87 - 50 - 37 cm 60 100 F=mag Styrofoam cup F:-mig Find force Find force F2of the F1 of the hanging hanging mass mass Fig 1: Experimental set up Step 6: Find Force on each end of the ruler by substituting vales for m, and m2 in equation F-mag and F=mag Step 7: Exchange weights m, and mą and repeat from step 4 Step 8: There is no step 8 Distance from Torque center of ruleri= Fr (1cm - 0.01m) Sum of F- mg (N) g- 9.8 m/s² Mass (m) in Kg Obs # Torques (1g-0.001kg) i= 11+ 12 0.05 r = m 1 F2 m2 0.10 : =-m F 0.10 r1 = m m2 0.05 12 = - m Is sum of torque zero?

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Note: 1. Mass in units of Kgs, length in unit of meters At equilibrium, the meter stick must be horizontal to ground. 3. This experiment should be performed on a flat surface. 2. Procedure: To find center of mass of the ruler Experimental setup: Part A: Prove Sum of Torques = 0 CHAR Step 1 Step 2 Step 3a Step 3 Step 1: Gently insert one end of ball bearing at the center of the Styrofoam cup as shown in step 1 Step 2: Apply tape at the center of the ruler followed by placing ruler on the ball bearing, such that the ruler is balanced. Step 3, 3a: Record center of mass, the point where the ruler is balanced is the center of mass of ruler, record the observation (indicated by arrow) The center of mass of the ruler = re = cm = CHARLYS LE Step 4 Step 4: Hang m, = 50g and m;=100g mass on the ruler with a string (see fig above) followed by gently moving m, or m, left or right on ruler, such that the meter stick is balanced as in step 2 Step 5: Find the distance ri and rz from re, record data in observation table below. Example: In figure below, r2 = - 20 + 50 = 30 cm, r = 87 - 50 = 37 cm 20 30 50 60 F=mg Styrofoam cup F=mig Find force Find force F1 of the hanging F2of the hanging mass mass Fig 1: Experimental set up Step 6: Find Force on each end of the ruler by substituting vales for m, and m2 in equation F;=mag and F2=m2g Step 7: Exchange weights m, and m2 and repeat from step 4 Step 8: There is no step 8 Distance from Torque T= Fr Sum of F= mg (N) g= 9.8 m/s? Mass (m) in Kg Obs. # center of ruler Torques (1g=0.001kg) (1cm = 0.01m) T= T1+ T2 F, T = F,n 0.05 "1 = m F2 = m2 0.10 12 =-m F1 0.10 2 m2 0.05 12 = - m Is sum of torque zero? What are the possible errors in the lab?
Transcribed Image Text:Note: 1. Mass in units of Kgs, length in unit of meters At equilibrium, the meter stick must be horizontal to ground. 3. This experiment should be performed on a flat surface. 2. Procedure: To find center of mass of the ruler Experimental setup: Part A: Prove Sum of Torques = 0 CHAR Step 1 Step 2 Step 3a Step 3 Step 1: Gently insert one end of ball bearing at the center of the Styrofoam cup as shown in step 1 Step 2: Apply tape at the center of the ruler followed by placing ruler on the ball bearing, such that the ruler is balanced. Step 3, 3a: Record center of mass, the point where the ruler is balanced is the center of mass of ruler, record the observation (indicated by arrow) The center of mass of the ruler = re = cm = CHARLYS LE Step 4 Step 4: Hang m, = 50g and m;=100g mass on the ruler with a string (see fig above) followed by gently moving m, or m, left or right on ruler, such that the meter stick is balanced as in step 2 Step 5: Find the distance ri and rz from re, record data in observation table below. Example: In figure below, r2 = - 20 + 50 = 30 cm, r = 87 - 50 = 37 cm 20 30 50 60 F=mg Styrofoam cup F=mig Find force Find force F1 of the hanging F2of the hanging mass mass Fig 1: Experimental set up Step 6: Find Force on each end of the ruler by substituting vales for m, and m2 in equation F;=mag and F2=m2g Step 7: Exchange weights m, and m2 and repeat from step 4 Step 8: There is no step 8 Distance from Torque T= Fr Sum of F= mg (N) g= 9.8 m/s? Mass (m) in Kg Obs. # center of ruler Torques (1g=0.001kg) (1cm = 0.01m) T= T1+ T2 F, T = F,n 0.05 "1 = m F2 = m2 0.10 12 =-m F1 0.10 2 m2 0.05 12 = - m Is sum of torque zero? What are the possible errors in the lab?
Theory: The rotational equivalent of force is defined as Torque(7). When a force Fis applied to a rigid body at any point away from the center of mass, a torque is produced. Torque, 1 (Greek letter, tau), can be defined as the tendency to cause rotation. The magnitude of the vector is: i = Frsine Where, r is the distance from the point of rotation to the point at which the force is being applied (i.e., lever arm), and Fr sin 0 is the component of the force perpendicular to r. The dimension of torque is [M] [L] [T*] with unit of N.m In part A, all forces will be acting normal (perpendicular) to the meter stick: 0 = 90°; therefore, sin 0 = 1. The equation for torque is simplified: i = Fr Equilibrium is reached when the net force and net torque on an object are zero. The first condition is that the vector sum of the forces must equal zero: ΣF-ΣΕ-Σ , -Σ 0 Newton The second condition that must be met is that the net torques about any axis of rotation must equal zero. We will use the standard convention for summing torques. Torques that tend to cause counterclockwise rotation(Tcc) will be positive torques, while torques that tend to cause clockwise rotation(Tc), will be negative torque. Στ= Στ. - Στ - Νm The system under consideration for this experiment will need to not only attain equilibrium, but also remain in equilibrium. This will require that the object be in stable equilibrium, meaning if a slight displacement of the system occurs, the system will return to its original position. If the system were to move farther from its original position when given a slight displacement, it would be in unstable and not be in equlibrium. Apparatus: 50 grams mass, 100 gram mass, Tape, 8oz Styrofoam cup, 30 cm Ruler, Thread
Transcribed Image Text:Theory: The rotational equivalent of force is defined as Torque(7). When a force Fis applied to a rigid body at any point away from the center of mass, a torque is produced. Torque, 1 (Greek letter, tau), can be defined as the tendency to cause rotation. The magnitude of the vector is: i = Frsine Where, r is the distance from the point of rotation to the point at which the force is being applied (i.e., lever arm), and Fr sin 0 is the component of the force perpendicular to r. The dimension of torque is [M] [L] [T*] with unit of N.m In part A, all forces will be acting normal (perpendicular) to the meter stick: 0 = 90°; therefore, sin 0 = 1. The equation for torque is simplified: i = Fr Equilibrium is reached when the net force and net torque on an object are zero. The first condition is that the vector sum of the forces must equal zero: ΣF-ΣΕ-Σ , -Σ 0 Newton The second condition that must be met is that the net torques about any axis of rotation must equal zero. We will use the standard convention for summing torques. Torques that tend to cause counterclockwise rotation(Tcc) will be positive torques, while torques that tend to cause clockwise rotation(Tc), will be negative torque. Στ= Στ. - Στ - Νm The system under consideration for this experiment will need to not only attain equilibrium, but also remain in equilibrium. This will require that the object be in stable equilibrium, meaning if a slight displacement of the system occurs, the system will return to its original position. If the system were to move farther from its original position when given a slight displacement, it would be in unstable and not be in equlibrium. Apparatus: 50 grams mass, 100 gram mass, Tape, 8oz Styrofoam cup, 30 cm Ruler, Thread
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