Experiment 6: Conservation of Mechanical Energy

The aim of the experiment is to test the principle of conservation of mechanical energy for motion under gravity. The equipment needed includes

  1. air table
  2. blower
  3. smart pulley
  4. vernier caliper
  5. puck
  6. hunger
  7. weights
  8. balance

Theory

The Smart Pulley

The smart pulley is essentially a photo gate that allows you to measure time intervals for consecutive constant distance intervals. One distance interval is one tenth of its circumference.

The acceleration may be calculated by using two consecutive intervals. Let t1 be the first time interval of the two consecutive intervals and t2 be the second time interval of the two consecutive time intervals. Let the constant distance interval be represented by s. If the acceleration, a, is constant, then the equations of a uniformly accelerated motion apply. For the first interval the following equation applies:

s = v1i t1 + at1 2 ⁄ 2 . . . . . .(1)

where v1i is the initial velocity of the first interval. And for the second interval the following equation applies:

s = v2i t2 + at2 2 ⁄ 2 . . . . . .(2)

where v2i is the initial velocity of the second interval. The final velocity, v1f, of the first interval is the same as the initial velocity, v2i, of the second interval. Therefore,

v2i = v1f = v1i + at1 . . . . . .(3)

Substituting equation (3) into equation (2) we obtain

s = v1it1t2 + a ( t1t2 + t1 2 ⁄ 2 ) . . . . . ( 4 )

The time intervals, t1 , t2 and the constant distance interval, s, will be measured. Thus, we will have only two unknowns, v1i and a, in equations (1) and (4). That means we can solve for the acceleration a simultaneously in terms of t1 , t2 and s to obtain

a = 2s ( t1 - t2 ) ⁄ [ t1 t2 ( t1 + t2 ) ] . . . . . ( 5 )

That is, if two consecutive time intervals and the constant distance interval are measured, the acceleration can be calculated.

Once the acceleration has been calculated, the initial velocity, v1i , of the first interval can be obtained from equation (1) as:

v1i = s ⁄ t1 - at1 ⁄ 2 . . . . . ( 6 )

Similarly the initial velocity of the second time interval, v2i , (or the final velocity of the first interval, v1f ) may be obtained from:

v2i = v1f = v1i + at1 ⁄ 2 . . . . . (7)

And substituting for v1i from equation (6) into equation (7) we get

v2i = v1f = s ⁄ t1 + at1 ⁄ 2 . . . . . ( 8 )

The final velocity of the second interval, v2f , may be calculated from

v2f = v2i + at2 . . . . . ( 9 )

And substituting for v2i from equation (8) into equation (9) we obtain

v2f = s ⁄ t1 + a ( t1 + 2t2 ) ⁄ 2 . . . . . ( 10 )

Conservation of Mechanical Energy

The principle of mechanical energy states that if all the forces with non-zero contribution to the work done acting on a system are conservative (forces for which the work done is path independent), then the mechanical energy of the system is conserved. Mechanical energy ( ME ) of a system is the sum of the kinetic energy ( KE ) and potential energy ( PE ) of the system.

ME = KE + PE . . . . . ( 11 )

The principle of conservation of mechanical energy may be mathematically written as

MEf = MEi

or

KEf + PEf = KEi + PEi . . . . . ( 12 )

Where MEi , KEi and PEi are the initial mechanical, kinetic and potential energies respectively and MEf , KEf and PEi are the final mechanical, kinetic and potential energies respectively.

The kinetic energy, KE, of an object of mass m and speed v is given by

KE = mv 2 ⁄ 2 . . . . . ( 13 )

And if the conservative force acting on the system is gravitational force, the potential energy, PEg , of an object of mass m and y-coordinate (vertical coordinate) y with respect to a certain reference point (choice of reference point is arbitrary as far as the same reference point is used consistently) is given by

PEg = m|g|y . . . . . ( 14 )

In this experiment, the system will involve an object of mass m1 sliding on an air table attached to a hanging object of mass m2 via a string. The force of friction acting on the sliding object will be reduced to a minimum by means of a blower and can be neglected. Thus, the only force with non-zero contribution to the work done acing on the system is the force of gravity acting on the hanging object (the normal force exerted by the surface of the table on the sliding object has no contribution to the work done because it is perpendicular to its trajectory). Therefore, since gravity is a conservative force, the mechanical energy of the system is conserved.

Both objects will be moving with the same speed, . Thus, the kinetic energy of the system is given by

KE = ( m1 + m2 ) v 2 ⁄ 2 . .. . . ( 15 )

If the y-coordinate of the sliding object is y1 and the y-coordinate of the hanging object is y2 , then the potential energy of the system is given by

PE = m1 |g|y1 + m2 |g|y2 . . . . . ( 16 )

Now the conservation of mechanical energy for this system may be obtained by substituting equations (15) and (16) into equation (12).

( m1 + m2 ) vf 2 ⁄ 2 + m1 |g|y1f + m2 |g|y2f = ( m1 + m2 ) vi 2 ⁄ 2 + m1 |g|y1i + m2 |g|y2i . . . . . ( 17 )

Since m2 is sliding in a horizontal surface, its y-coordinate will not change; that is y1f = y1i and equation ( 17 ) reduces to

( m1 + m2 ) vf 2 ⁄ 2 + m2 |g|y2f = ( m1 + m2 ) vi 2 ⁄ 2 + m2 |g|y2i . . . . . ( 18 )

or

{ ( m1 + m2 ) vf 2 ⁄ 2 + m2 |g|y2f } ⁄ { ( m1 + m2 ) vi 2 ⁄ 2 + m2 |g|y2i } = 1 . . . . . ( 19 )

Procedure

  1. To test the principle of conservation of mechanical energy for motion under gravity.

    1. Measure the mass ( m1 ) of the puck (cylindrical object) by means of a balance.

      puck mass ( m1 ) = kg = 0.038 kg

      Record this in the first row of the first column of the Table.

    2. Measure the diameter of the smart pulley by means of a vernier caliper. (Note: make sure to measure the diameter between the edges where the string would slide)

      diameter ( d ) =

      The circumference of the pulley is divided into ten equal distance intervals. The smart pulley measures the time taken for each of these intervals. To determine the length ( s ) of this interval, first calculate the circumference and then divide by ten.

      interval distance ( s ) = πd ⁄ 10 = m

    3. Connect the blower to the air table. Place the puck on the air table and turn on the blower. Level the table by adjusting the screws (The legs of the air table are adjustable screws) until there is no net tendency for the puck to go in any direction.

    4. Turn off the blower. Connect the smart pulley to the air table along the edge of the table.

    5. Connect the puck (cylindrical plastic) to a string, pass the string through the smart pulley and then connect it to the hanger. The mass ( m2 ) of the hanger is marked on it. Read this mass.

      hanger mass ( m2 ) = kg = 0.005 kg

    6. connect the smart pulley to the computer (on the computer choose smart pulley and then motion timer).

    7. Position the puck at the rear of the air table and let it stand by itself (friction will stop it from sliding).

    8. Turn on the blower to remove friction and the puck will start sliding.

    9. The computer will record time taken for the constant distance intervals of length s. (i.e, one tenth of the circumference of the pulley.) Pick any two consecutive time intervals, preferably the third and the fourth time intervals. Represent the first of these time intervals by t1 and the second one by t2 .

      first time interval ( t1 ) = s

      second time interval ( t2 ) = s

      Record t1 on the first row of column 2 and t2 on the first row of column 3 of the Table .

    10. Applying the equations of a uniformly accelerated motion to the two intervals and solving the system of simultaneous equations, we can obtain the following expression for the acceleration ( a ) of the system in terms of the time intervals t1 and t2 , and the distance traveled in each interval, s (For details look at the theory section of the manual).

      a = 2s ( t1 - t2 ) ⁄ [ t1 t2 ( t1 + t2 ) ]

      Using the time intervals, t1 and t2 obtained in procedure (9.) and the distance of one interval ( s ) obtained from procedure (2.), calculate the acceleration of the system.

      acceleration ( a ) = 2s ( t1 - t2 ) ⁄ [ t1 t2 ( t1 + t2 ) ] = m ⁄ s 2

      Record this on the first row of column 4 of the table.

    11. Since friction has been removed the only force with non-zero contribution to the work done on the system is the force of gravity acting on the hanger which is conservative. Therefore we expect the mechanical energy of the system to remain constant. To test this we will take our initial event to be the beginning of the first time interval, t1 and our final event to be the end of the second time interval, t2. As shown in the theory section, theinitial speed is given as follows:

      vi = s ⁄ t1 - at1 ⁄ 2

      Using the constant distance interval s obtained in procedure (2.), the acceleration obtained in procedure (10.), and the first of the two consecutive time intervals t1 obtained in procedure (9.), calculate the initial speed vi .

      initial speed ( vi ) = s ⁄ t1 - at1 ⁄ 2 = m ⁄ s

      Record this at the first row of column 5 of the table.

      The final speed ( vf ), which is equal to the final speed of the second time interval ( v2f ), is given as follows as shown in the theory section:

      vf = s ⁄ t1 + ( a ⁄ 2 ) ( t1 + 2t2 )

      Using the constant distance interval s obtained in procedure (2.), the acceleration a obtained in procedure (10.), and the consecutive time intervals t1 and t2 , calculate the final speed vf.

      final speed ( vf ) = s ⁄ t1 + ( a ⁄ 2 ) ( t1 + 2t2 ) = m ⁄ s

      Record this at the first row of column 6 of the table.

      The mechanical energy of the system is the sum of the kinetic energy ang gravitational potential energy of the system. The sliding object is moving in a horizontal surface and thus its potential energy remains the same and need not be considered (look at the theory section for details). If we choose our reference point to be the initial location of the hanger ( m2 ), then the initial potential energy of the hanger is zero. Therefore, with this choice of reference point, the initial mechanical energy ( MEi ) involves only the kinetic energy of the system.

      MEi = ( m1 + m2 ) vi 2 ⁄ 2

      Calculate the initial mechanical energy.

      initial mechanical energy ( MEi ) = ( m1 + m2 ) vi 2 ⁄ 2 = J

      Record this at the first row of column 7 of the table

      Since the distance traveled in one interval is equal to s, during the two consecutive time intervals, the location of the hanger ( m2 ) will be lowered by a distance of 2s. And thus, its final y-coordinate with respect to its location is equal to -2s (negative because it is going down). Therefore the potential energy at the final location is m2 |g| ( -2s ) and the final mechanical energy ( MEf ) is given by

      MEf = ( m1 + m2 ) vf 2 ⁄ 2 - m2 |g| ( 2s )

      Calculate the final mechanical energy of the system.

      final mechanical energy ( MEf ) = ( m1 + m2 ) vf 2 ⁄ 2 - m2 |g| ( 2s ) = J

      Record this at the first row of column 8 of the table.

    12. Calculate the ratio between the final mechanical energy (column 8) and the initial mechanical energy (column 7).

      ratio = MEf ⁄ MEi =

      Record this at the first row of column 9 of the table.

    13. Repeat procedures (7.) through (12.) by attaching an additional mass of 10 g, 20 g, 30 g and 40 g on the puck to complete the table. (Tape the additional mass on the puck so that it doesn’t move around on top of the puck) Note: the value of m1 is the sum of the mass of the puck (procedure 1.) and the additional mass.

      Table 1
      update table
    14. Calculate the average of the numerical values of the ratio between final and initial mechanical energy obtained on column 9 of the table.

      average =

    15. The ratio the final mechanical energy and the initial mechanical energy is expected to be one because mechanical energy is conserved. Compare the average of the ratio between final and initial mechanica energy obtained in procedure (13.) with the expected value of one by calculating the percentage error.

      % error = |( average ratio(14.) - 1 ) ⁄ 1| * 100% = %