Using the second condition of equilibrium (expressing torques about the center of the stick) find an...
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Using the second condition of equilibrium (expressing
torques about the center of the stick) find an...
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...
the experimental part for this question is all answered there is only the last part where it says: Compute the net torque about 20 cm mark. Mass of the stick : 0.1383 kg Compu...
the experimental part for this question is all answered there is
only the last part where it says:
Compute the net torque about 20 cm
mark.
Mass of the stick : 0.1383 kg
Compute the net torque about the end of the stick.
please be sure to put detailed answer for both parts 1 and 2
thanks for the help!
Q1) Torques and equilibrium Part A: use the following methods to find the center of mass for a measuring stick...
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...
Suspend a mass m, =100g at or near the zero end of the meter stick. Move...
Suspend a mass m, =100g at or near the zero end of the meter stick. Move the meter stick in the support clamp until the meter stick is in equilibrium. Record this new equilibrium position as x," Using the total mass of the meter stick, calculate the clockwise and counterclockwise torques, and then calculate a percent difference. In this calculation, you will include the mass of the meter stick as if it were concentrated at its center of mass, x,...
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)...
The torques acting on the crane boom can be computed about an axis passing through the...
The torques acting on the crane boom can be computed about an axis passing through the top of the boom. (This is the place where the spring scale and mass hanger connect to the boom.) Apply the second condition for equilibrium to the crane boom and derive the resulting equation for the sum of the torques about this new point.
Period Date Name 39 Solitary Seesaw Purpose To identify the forces, lever arms, and torques for a system in rota...
Period Date Name 39 Solitary Seesaw Purpose To identify the forces, lever arms, and torques for a system in rotational 2 knife-edge lever clamps set of slotted masses 2 mass hangers fulcrum string or masking tape Gravity pulls on every part of an object. The average position of these pulls (the weight) is the center of gavity (CG) of the object. The sum of all these pulls is the weight of the object. The entire weight of the object is...
Torque Lab help I need to find the Mass of the meter stick (mms) using the...
Torque Lab help I need to find the Mass of the meter stick (mms) using the equation (M1+mmc)g(Xo-X1)=(M2+mmc)*g*(X2-Xo)+mms*g*(Xcm-Xo) xo= 03.m mmc= 0.01516kg m1=0.3kg x1=0.01m m2= 0.470 x2= 0.4m xcm= 0.495m
For equilibrium the clockwise rotation or torque must equal the counterclockwise! If this were not true...
For equilibrium the clockwise rotation or torque must equal the counterclockwise! If this were not true the object would start spinning. The torque of W about X is just W * Lw and the torque of T2 about X is T2 * L2. So for rotational equilibrium W * Lw = T2 * L2. The distances Lw and L2 are called the lever anus of the forces. 1. Using the preceding picture assume that the stick is a meter stick,...
A metal disc spins with an angular velocity ω0 about its center. A second disc (initially...
A metal disc spins with an angular velocity ω0 about its center. A second disc (initially stationary) is dropped on top of the first disc. The second disc is made of the same material as the first but has only half the diameter (so its mass will also be smaller). (a) Show that the moment of inertia of the second disc is 1/16th the moment of inertia of the first. (b) State why the angular momentum of the system (the...
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...
the experimental part for this question is all answered there is
only the last part where it says:
Compute the net torque about 20 cm
mark.
Mass of the stick : 0.1383 kg
Compute the net torque about the end of the stick.
please be sure to put detailed answer for both parts 1 and 2
thanks for the help!
Q1) Torques and equilibrium Part A: use the following methods to find the center of mass for a measuring stick...
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...
Suspend a mass m, =100g at or near the zero end of the meter stick. Move the meter stick in the support clamp until the meter stick is in equilibrium. Record this new equilibrium position as x," Using the total mass of the meter stick, calculate the clockwise and counterclockwise torques, and then calculate a percent difference. In this calculation, you will include the mass of the meter stick as if it were concentrated at its center of mass, x,...
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)...
The torques acting on the crane boom can be computed about an axis passing through the top of the boom. (This is the place where the spring scale and mass hanger connect to the boom.) Apply the second condition for equilibrium to the crane boom and derive the resulting equation for the sum of the torques about this new point.
Period Date Name 39 Solitary Seesaw Purpose To identify the forces, lever arms, and torques for a system in rotational 2 knife-edge lever clamps set of slotted masses 2 mass hangers fulcrum string or masking tape Gravity pulls on every part of an object. The average position of these pulls (the weight) is the center of gavity (CG) of the object. The sum of all these pulls is the weight of the object. The entire weight of the object is...
For equilibrium the clockwise rotation or torque must equal the counterclockwise! If this were not true the object would start spinning. The torque of W about X is just W * Lw and the torque of T2 about X is T2 * L2. So for rotational equilibrium W * Lw = T2 * L2. The distances Lw and L2 are called the lever anus of the forces. 1. Using the preceding picture assume that the stick is a meter stick,...
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