Experiment 4: Newton's Second Law

The aim of the experiment is to determine the value of gravitational acceleration by using Newton’s second law. 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.

Newton's Second Law

Newton’s secnd law states that the force ( F ) acting on an object is directly proportional to the acceleration ( a ) produced where the constant of proportionality is defined to be the mass ( m ) of the object.

F = ma

To demonstrate this relationship, we will use an arrangement where an object of mass m1 sliding on a table is attached to a hanging object of mass m2 by means of a string passing through a pulley. The friction between the sliding object ( m1 ) and the table will be minimized with the help of an air table and can be neglected. Therefore, with friction neglected the only force acting on the system is the weight ( W ) of the hanging object ( m2 ).

F = W = m2 |g|

, where g is the gravitational acceleration whose value is –9.8 m ⁄ s 2. Since both masses are being accelerated by this force, the mass that goes to Newton’s second law should be the sum of the masses of both objects. Thus, if the acceleration of the system is a , then from Newton’s second law we have

m2 |g| = ( m1 + m2 ) a

Therefore if the masses m1 , m2 and the acceleration, a, are measured, the value of gravitational acceleration can be calculated.

|g| = ( m1 + m2 ) a ⁄ m2

Procedure

  1. To determine the value of gravitational acceleration by using Newton’s second law.

    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 ) = m = 0.049 m

      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 = 0.046 s

      second time interval ( t2 ) = s = 0.040 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 the force responsible for the motion is the weight of the hanging object (W = m2 ) and both masses are being accelerated, applying Newton’s second law we get the following equation:

      m2 |g| = ( m1 + m2 ) a

      Therefore if the masses and their acceleration are known gravitational acceleration, |g|, may be calculated from

      |g| = ( m1 + m2 ) a ⁄ m2

      Using the mass, m1 , obtained from procedure (1.), the mass, m2 , obtained from procedure (5.), and the acceleration of the system obtained from procedure (10.), calculate the numerical value of gravitational acceleration,

      gravitational acceleration ( |g| ) = ( m1 + m2 ) a ⁄ m2 = m ⁄ s 2

      Record this on the first row of column 5 of the Table.

    12. Repeat procedure (7.) through (11.) by adding an additional mass of 10 g, 20 g, 30 g and 40 g to the puck to complete the table. (Note: the mass m1 will be the sum of the mass of the puck and the additional mass) Attach the additional mass to the puck with a tape to make sure it doesn’t move around on the top of the puck.

      Table 1
      update table
    13. Calculate the average of the numerical values of the gravitational acceleration obtained on column 5 of the table.

      average = m ⁄ s 2

    14. Compare the average of the gravitational accelerations obtained in procedure (13.) with the known value of gravitational acceleration of 9.8 m ⁄ s 2 by calculating the percentage error.

      % error = |(gravitational acceleration (13.) - 9.8 m ⁄ s 2 ) ⁄ ( 9.8 m ⁄ s 2 )| * 100% = %