Experiment 8: Torque

1. A meter stick
2. hangers
3. a balance
4. a set of weights.

Theory

Torque

τ = Fr_⊥

If the angle formed between the line joining the point of rotation to the point of application of the force and the line of action of the force is θ ,then r_⊥ may be expressed interms of this angle and the distance ( r ) between the point of rotation and the point of application of force as r_⊥ = r sin ( θ ). Therefore we may also write the following expression for the torque:

τ = Fr sin ( θ )

A torque due to a given force is taken to be positive if the effect of the force is to cause a counterclockwise rotation about the point of rotation, while it is taken to be negative if its effect is to cause a clockwise rotation.

There are two kinds of equilibrium: translational and rotational equilibrium. An object is said to be in translational equilibrium if it is either at rest or is moving in a straight line with a constant speed. An object will be in translational equilibrium if the net force acting on the object is zero. The condition of translational equilibrium may be written in component form as

ΣF_x = F_1x + F_2x + . . . = 0

ΣF_y = F_1y + F_2y + . . . = 0

An object is said to be in rotational equilibrium if either it is at rest or rotating with a constant angular speed. An object will be in rotational equilibrium if the net torque acting on it is zero.

Στ = τ₁ + τ₂ + . . . = 0

Procedure

1. To demonstrate the condition of rotational equilibrium

   1. Measure the masses of the three hangers by means of a balance.

      hanger 1 mass ( m₁ ) = kg = 0.15 kg
      
      hanger 2 mass ( m₂ ) = kg = 0.2 kg
      
      hanger 3 mass ( m₃ ) = kg = 0.3 kg

   2. Put the meter stick into a pivot. (You can use the hanger without the suspending wires as a pivot).

   3. Suspend the pivot-meter stick system on the stand. Balance this system to find the center of gravity (that is the cm mark the pivot is located) of the system ( x_cg ).

      center of gravity ( x_cg ) = m

   4. Now consider the following mathematical problem: if hanger 1 ( m₁ ) is placed at the x₁ = 40 cm = 0.4 m mark and hanger 2 ( m₂ ) is placed at the x₂ = 10 cm = 0.1 m mark on the meter stick, where should the third hanger be placed if the ruler is to be balanced or be in equilibrium?

      This problem may be solved by using the condition of rotational equilibrium which states that for an object in rotational equilibrium the net torque should be zero. Taking torques about the pivot ( x_cg ) which is also the center of gravity. Note: The torque due to the weight of the ruler is zero since the torque is being taken about the center of gravity.)

      τ₁ + τ₂ + τ₃ = 0

      Torque ( τ ) about the pivot is obtained as the product of the magnitude of the force ( F ) and the perpendicular distance ( r_⊥ ) between the pivot and the line of action of the force. Torque is taken to be positive if the effect of the force is to produce counterclockwise rotation and is taken to be negative if its effect is to produce clockwise rotation.

      τ = ± Fr_⊥

      The force in this case is the weight of the hangers: F₁ = m₁|g|, F₂ = m₂|g|, and F₃ = m₃|g|. The perpendicular distance is the absolute value of difference between the positions of the hanger and the pivot.

      hanger 1 distance ( r_⊥1 ) = |x₁ - x_cg | = m
      
      hanger 2 distance ( r_⊥2 ) = |x₂ - x_cg | = m

      Applying the condition for rotational equilibrium, the following equation is obtained.

      m₁ |g|r_⊥1 + m₂ |g|r_⊥2 = m₃ |g|r_⊥3

      Solving for r_⊥3

      r_⊥3 = ( m₁ r_⊥1 + m₂ r_⊥2 ) ⁄ m₃

      Using the values of m₁ , m₂ and m₃ from procedure (1.), r_⊥1 and r_⊥2 from procedure (4.), solve for the perpendicular distance between hanger 3 and the pivot, |r_⊥3 .

      calculated hanger 3 distance ( r_⊥3 ) = ( m₁ r_⊥1 + m₂ r_⊥2 ) ⁄ m₃ = m

   Now to determine the location of hanger 3 experimentally, place hanger 1 at the 40 cm location and hanger 2 at the 20 cm location and find experimentally the location of hanger 3 that will balance the ruler. Determine the perpendicular distance ( r_⊥3 ) between hanger 3 and the pivot.

   measured hanger 3 distance ( r_⊥3 ) = m

   Compare the calculated and measured values of the distance between pivot and location of hanger 3 ( |x₃ - x_cg | ) by calculating the percentage error.

   % error = |{ calculated distance (4.) - measured distance (5.) } ⁄ { calculated distance (4.) }| * 100% = %

2. Put a mass of 150 g to hanger 1, a mass of200 g to hanger 2, and a mass of 250 g to hanger 3; and repeat procedures (2.) through (6.) to obtain the following values.

   1. Record the masses. Remember now the mass is the combination of the mass of the hanger and the additional mass.

      hanging mass 1 ( m₁ ) = kg
      
      hanging mass 2 ( m₂ ) = kg
      
      hanging mass 3 ( m₃ ) = kg

   2. Calculate the perpendicular distance between hanger 3 and the pivot that balances the meter stick.

      calculated hanger 3 distance ( r_⊥3 ) = ( m₁ r_⊥1 + m₂ r_⊥2 ) ⁄ m₃ = m

   Find the location of hanger 3 that balances the meter stick experimentally and determine the distance between hanger 3 and the pivot.

   measured hanger 3 distance ( r_⊥3 ) = m = 0.28 m

   Compare the calculated (procedure 2.) and experimental (procedure 3.) perpendicular distance between hanger 3 and the pivot by calculating the percentage error.

   % error = |{ calculated distance (2.) - measured distance (3.) } ⁄ { measured distance (3.) }| * 100% = %

3. To determine the mass of a meter stick by means of condition of rotational equilibrium and by means of a balance and then compare

   1. Change the location of the pivot ( x_o ) to the 60 cm mark of the meter stick.

   2. Put a mass of 150 g to hanger 1, a mass of 200 g to hanger 2, and a mass of 500 g to hanger 3.

      hanging mass 1 ( m₁ ) = kg = 0.15 kg
      
      hanging mass 2 ( m₂ ) = kg = 0.2 kg
      
      hanging mass 3 ( m₃ ) = kg = 0.5 kg

   3. Place hanger 1 on the x₁ = 40 cm mark and hanger 2 on the x₂ = 20 cm mark. Determine the distances between these hanger locations and the pivot. Remember the location of the pivot has changed.

      hanger 1 distance ( r_⊥1 ) = m
      
      hanger 2 distance ( r_⊥2 ) = m

   4. Balance the table and determine the location of the third hanger that balances the meter stick. Determine the distance between hanger 3 and the pivot.

      hanger 3 distance ( r_⊥3 ) = m

   5. The weight of the lever can be assumed to act at the center of gravity of the meter stick, x_cg, determined in procedure I.3. Determine the distance between the center of gravity of the ruler and the pivot.

      center of gravity distance ( r_⊥cg ) = x_o - x_cg = m

   6. The mass of the ruler ( m_r ) may be calculated by applying the condition of rotational equilibrium. From condition of rotational equilibrium, the torque due to the weight at the third location should balance the combined torques due to the weights at the first and second locations and the weight of the meter stick acting at the center of gravity of the ruler.

      m₃ r_⊥3 = m₁ r_⊥1 + m₂ r_⊥2 + m_r r_⊥cg

      Solving for the mass of the ruler ( m_r ):

      m_r = { m₃ r_⊥3 - m₁ r_⊥1 - m₂ r_⊥2 } ⁄ r_⊥cg

      Using the masses obtained in procedure (2.) and the perpendicular distances obtained in procedures (3.), (4.) and (5.), use this equation to calculate the mass of the meter stick.

      calculated meter stick mass ( m_r ) = { m₃ r_⊥3 - m₁ r_⊥1 - m₂ r_⊥2 } ⁄ r_⊥cg = kg

   7. Measure the mass of the meter stick by means of a balance.

      measured meter stick mass ( m_r ) = kg = 0.15 kg

   8. Compare the calculated mass (procedure 6.) and measured mass (procedure 7.) of the meter stick by calculating the percentage error.

      % error = |{measured mass (6.) - calculated mass (5.) } ⁄ { measured mass (6.)} = %