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Suspend a mass m, =100g at or near the zero end of the meter stick. Move...
at With the meter stick on the support stand, suspend a gram mass m = 100g...
at With the meter stick on the support stand, suspend a gram mass m = 100g the 15-cm mark on the meter stick. Then adjust the lever arm for a gram mass m2 = 200g on the other side of the axis. See Figure. X axis X X2 m m2 Record the mass and position x as read on the meter stick and then record the lever arms. Compute the torques and find a percent difference between the clockwise (Tow)...
Place m, = 100g at the 20-cm mark and m2 = 200g is at the 60-cm...
Place m, = 100g at the 20-cm mark and m2 = 200g is at the 60-cm mark on the meter stick. Experimentally determine the position for my = 50g so that the system is in equilibrium. Follow a procedure similar to steps 2 and 3. Compute the percent difference of the clockwise and counterclockwise torques. See Figure. X. axis X X2 X₂ m. m2 m. Case 2 m = 100 g x1 = 20 cm TE 200 g ra T...
Place an unknown mass at the 10 cm mark of the meter stick. Suspend from the...
Place an unknown mass at the 10 cm mark of the meter stick. Suspend from the other side a counter mass m2 = 300g and adjust its position until the system is in static equilibrium. Using, Στ = 0 calculate the unknown mass m². Remove the unknown mass and determine its mass on the laboratory balance. See Figure. This is the accepted mass. Calculate % error. % Error lexperimentalacceptedy 100 accepted X axis X X2 m m, m=? X =...
Procedure 1, 2, 3 Mass of a clamp: 21.5 True mass of meter stick: _133.6 g...
Procedure 1, 2, 3 Mass of a clamp: 21.5 True mass of meter stick: _133.6 g Center of mass of meter stick: _50 cm True weight of unknown mass: _212.99_ Procedure 4 Position of the 100-g mass: 10.0 cm Position of equilibrium: _30.9 cm Mass of the stick from method of torques: Procedure 5 Position of the 100-g mass: 10.0 cm Position of the 200-g mass: _90.0 cm Position of equilibrium of meter stick: Procedure 6 Position of unknown mass...
can ignore friction when they ispherical bowl of II. A) In this part of the laboratory...
can ignore friction when they ispherical bowl of II. A) In this part of the laboratory you will explore the basic concept of rotational equilibrium, i.e. the balancing of torques about a fulcrum or pivot point. You will be given values for 3 masses along with given positions (X) on a meter stick for which two of these masses will be hung. Your task is to determine the position that the third mass must be hung in order to balance...
Torques and Center of Mass. The Experiment: In this experiment, you balance a meter stick, to...
Torques and Center of Mass. The
Experiment:
In this experiment, you balance a meter stick, to balance the
meter stick, attach masses at positions until the system is in
equilibrium.
The meter stick acts as if all its mass was concentrated at its
center of mass. With the fulcrum at the center of mass, r (the
distance from the axis of rotation to the place where the force is
applied) is 0, so there’s no torque due to the meter...
A uniform wooden meter stick has a mass of m = 799 g. A clamp can be attached to the measuring stick at any point Palong the stick so that the stuck can rotate freely about point P
A uniform wooden meter stick has a mass of m = 799 g. A clamp can be attached to the measuring stick at any point Palong the stick so that the stuck can rotate freely about point P, which is at a distance d from the zero-end of the stick as shown. Part (a) Calculate the moment of inertia in kg-m of the meter stick if the pivot point P is at the 50-cm mark. Part (b) Calculate the moment of inertia...
0/30 Torque and Conditions for Equilibrium. The pivot point of a meter stick is at 50.0...
0/30 Torque and Conditions for Equilibrium. The pivot point of a meter stick is at 50.0 cm. The first mass of 35.0 kg is placed at the 23.0 cm mark of the meterstick. What is the location of a mass - 65.0 kg that balances the meter stick (Sum of torques
Place m = 100g at the 20-cm mark and m2 = 200g is at the 60-cm...
Place m = 100g at the 20-cm mark and m2 = 200g is at the 60-cm mark on the meter stick. Experimentally determine the position for mz = 50g so that the system is in equilibrium. Follow a procedure similar to steps 2 and 3. Compute the percent difference of the clockwise and counterclockwise torques. See Figure. X axis X X2 X₂ ma m2 m3 Case 2 x1 = 20 cm CC m = 100 g m2 = 200 g...
A54.0 meter stick is balanced at its midpoint (50.0 cm, zero point is a left end...
A54.0 meter stick is balanced at its midpoint (50.0 cm, zero point is a left end of stick). Then 2500 welt hung with a light string from the 1967.0 cm point, and a 125 g weight is hung from the 16,0 cm point the figure below). Calculate the clockwise and counterclockwise torques acting on the board due to the four forces shown about the following axes: 50 cm 1) Calculate the clockwise torque if the axis is the 0 cm...
at With the meter stick on the support stand, suspend a gram mass m = 100g the 15-cm mark on the meter stick. Then adjust the lever arm for a gram mass m2 = 200g on the other side of the axis. See Figure. X axis X X2 m m2 Record the mass and position x as read on the meter stick and then record the lever arms. Compute the torques and find a percent difference between the clockwise (Tow)...
Place m, = 100g at the 20-cm mark and m2 = 200g is at the 60-cm mark on the meter stick. Experimentally determine the position for my = 50g so that the system is in equilibrium. Follow a procedure similar to steps 2 and 3. Compute the percent difference of the clockwise and counterclockwise torques. See Figure. X. axis X X2 X₂ m. m2 m. Case 2 m = 100 g x1 = 20 cm TE 200 g ra T...
Place an unknown mass at the 10 cm mark of the meter stick. Suspend from the other side a counter mass m2 = 300g and adjust its position until the system is in static equilibrium. Using, Στ = 0 calculate the unknown mass m². Remove the unknown mass and determine its mass on the laboratory balance. See Figure. This is the accepted mass. Calculate % error. % Error lexperimentalacceptedy 100 accepted X axis X X2 m m, m=? X =...
Procedure 1, 2, 3 Mass of a clamp: 21.5 True mass of meter stick: _133.6 g Center of mass of meter stick: _50 cm True weight of unknown mass: _212.99_ Procedure 4 Position of the 100-g mass: 10.0 cm Position of equilibrium: _30.9 cm Mass of the stick from method of torques: Procedure 5 Position of the 100-g mass: 10.0 cm Position of the 200-g mass: _90.0 cm Position of equilibrium of meter stick: Procedure 6 Position of unknown mass...
can ignore friction when they ispherical bowl of II. A) In this part of the laboratory you will explore the basic concept of rotational equilibrium, i.e. the balancing of torques about a fulcrum or pivot point. You will be given values for 3 masses along with given positions (X) on a meter stick for which two of these masses will be hung. Your task is to determine the position that the third mass must be hung in order to balance...
Torques and Center of Mass. The
Experiment:
In this experiment, you balance a meter stick, to balance the
meter stick, attach masses at positions until the system is in
equilibrium.
The meter stick acts as if all its mass was concentrated at its
center of mass. With the fulcrum at the center of mass, r (the
distance from the axis of rotation to the place where the force is
applied) is 0, so there’s no torque due to the meter...
0/30 Torque and Conditions for Equilibrium. The pivot point of a meter stick is at 50.0 cm. The first mass of 35.0 kg is placed at the 23.0 cm mark of the meterstick. What is the location of a mass - 65.0 kg that balances the meter stick (Sum of torques
Place m = 100g at the 20-cm mark and m2 = 200g is at the 60-cm mark on the meter stick. Experimentally determine the position for mz = 50g so that the system is in equilibrium. Follow a procedure similar to steps 2 and 3. Compute the percent difference of the clockwise and counterclockwise torques. See Figure. X axis X X2 X₂ ma m2 m3 Case 2 x1 = 20 cm CC m = 100 g m2 = 200 g...
A54.0 meter stick is balanced at its midpoint (50.0 cm, zero point is a left end of stick). Then 2500 welt hung with a light string from the 1967.0 cm point, and a 125 g weight is hung from the 16,0 cm point the figure below). Calculate the clockwise and counterclockwise torques acting on the board due to the four forces shown about the following axes: 50 cm 1) Calculate the clockwise torque if the axis is the 0 cm...
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