Practical work moving a body in a circle. Determination of body weight by weighing on a scale. Circulation time, s
For 9th grade (I.K.Kikoin, A.K.Kikoin, 1999),
task №5
to the chapter " LABORATORY WORKS».
Purpose of the work: to make sure that when a body moves in a circle under the action of several forces, their resultant is equal to the product of the body mass and acceleration: F = ma. For this, a conical pendulum is used (Fig. 178, a).
On a body attached to a thread (in work this is a load of
set in mechanics) the force of gravity F 1 and the elastic force F 2 act. Their resultant is equal to
Force F imparts centripetal acceleration to the load
(r is the radius of the circle along which the load moves, T is the period of its revolution).
To find the period, it is convenient to measure the time t of a certain number N of revolutions. Then T =
The modulus of the resultant F of the forces F 1 and F 2 can be measured by compensating for it with the elastic force F of the control spring of the dynamometer as shown in Figure 178, b.
According to Newton's second law,
When substituting into
this is the equality of the experimentally obtained values F ynp , m and a it may turn out that the left side of this equality differs from unity. This allows us to estimate the error of the experiment.
Measuring tools: 1) ruler with millimeter divisions; 2) a clock with a second hand; 3) dynamometer.
Materials: 1) tripod with coupling and ring; 2) strong thread; 3) a sheet of paper with a drawn circle with a radius of 15 cm; 4) weight from the mechanics set.
Work order
1. Tie a thread about 45 cm long to a weight and hang it from the tripod ring.
2. One of the students grab the thread at the point of suspension with two fingers and rotate the pendulum.
3. For the second student, use a tape to measure the radius r of the circle along which the load moves. (You can draw a circle in advance on paper and set the pendulum in motion along this circle.)
4. Determine the period T of revolution of the pendulum using a clock with a second hand.
To do this, the student, rotating the pendulum, in time with its revolutions, says out loud: zero, zero, etc. The second student with a watch in his hands, having caught the convenient moment in the second hand to start counting, says: “zero,” after which the first student out loud counts the number of revolutions. After counting 30-40 revolutions, the time interval t is recorded. The experiment is repeated five times.
5. Calculate the average acceleration value using formula (1), taking into account that with a relative error of no more than 0.015 we can assume π 2 = 10.
6. Measure the modulus of the resultant F, balancing it with the elastic force of the dynamometer spring (see Fig. 178, b).
7. Enter the measurement results in the table:
8. Compare attitude
with unity and draw a conclusion about the error in the experimental verification of what centripetal acceleration imparts to the body vector sum forces acting on it.
The weight from the mechanics set, suspended on a thread fixed at the top point, moves in horizontal plane along a circle of radius r under the action of two forces:
gravity
and elastic force N.
The resultant of these two forces F is directed horizontally towards the center of the circle and imparts centripetal acceleration to the load.
T is the period of circulation of the load in a circle. It can be calculated by calculating the time during which the load makes a certain number of full revolutions
Let's calculate centripetal acceleration using the formula
Now, if you take a dynamometer and attach it to a load, as shown in the figure, you can determine the force F (the resultant of the forces mg and N.
If the load is deflected from the vertical by a distance r, as when moving in a circle, then the force F is equal to the force that caused the load to move in a circle. We get the opportunity to compare the value of the force F obtained by direct measurement and the force ma calculated from the results of indirect measurements and
compare attitude
with one. In order for the radius of the circle along which the load moves to change more slowly due to the influence of air resistance and this change to have a slight effect on the measurements, it should be chosen small (about 0.05 ~ 0.1 m).
Completing of the work
Computations
Error estimation. Measurement accuracy: ruler -
stopwatch
dynamometer
Let's calculate the error in determining the period (assuming that the number n is determined exactly):
We calculate the error in determining acceleration as:
Determination error ma
(7%), that is
On the other hand, we measured the force F with the following error:
This measurement error is, of course, very large. Measurements with such errors are only suitable for rough estimates. This shows that the deviation ratio
from one can be significant when using the measurement methods we used *.
1 * So you shouldn't be embarrassed if this lab involves
will be different from unity. Just carefully evaluate all measurement errors and draw the appropriate conclusion.
Subject: Study of the movement of a body in a circle.
Goal of the work: determination of the centripetal acceleration of the ball when it uniform motion around the circumference.
Equipment:
- tripod with coupling and foot;
- measuring tape;
- compass;
- laboratory dynamometer;
- scales with weights;
- ball on a string;
- a piece of cork with a hole;
- paper;
- ruler.
Theoretical part
Experiments are carried out with a conical pendulum. A small ball moves in a circle with radius R. In this case the thread AB, to which the ball is attached, describes the surface of a right circular cone. There are two forces acting on the ball: gravity mg and thread tension F(see fig A). They create a centripetal acceleration a n, directed radially towards the center of the circle. The acceleration modulus can be determined kinematically. It is equal to:
a n = ω 2 R = 4π 2 R/T 2
To determine acceleration, you need to measure the radius of the circle R and the period of revolution of the ball in a circle T. Centripetal (normal) acceleration can also be determined using the laws of dynamics. According to Newton's second law ma = mg + F. Let's break down the force F into components F 1 And F 2, directed radially to the center of the circle and vertically upward. Then Newton's second law can be written as follows:
ma = mg + F 1 + F 2.
We choose the direction of the coordinate axes as shown in the figure b. In projection onto the O 1 Y axis, the equation of motion of the ball will take the form: 0 = F 2 - mg. From here F2 = mg. Component F 2 balances gravity mg, acting on the ball. Let's write Newton's second law in projection onto the axis O 1 X: ma n = F 1. From here and n = F 1 /m. Component modulus F 1 can be determined different ways. Firstly, this can be done using the similarity of triangles OAV And FBF 1:
F 1 /R = mg/h
From here F 1 = mgR/h And a n = gR/h.
Secondly, the modulus of the component F 1 can be directly measured with a dynamometer. To do this, we pull the ball with a horizontal dynamometer to a distance equal to the radius R circles (Fig. V), and determine the dynamometer reading. In this case, the elastic force of the spring balances the component F 1. Let us compare all three expressions for and n:
a n = 4π 2 R/T 2 , a n = gR/h, a n = F 1 /m
and make sure that the numerical values of centripetal acceleration obtained by three methods are close to each other.
In this work, time should be measured with the greatest care. For this purpose it is useful to count down perhaps larger number N revolutions of the pendulum, thereby reducing the relative error.
There is no need to weigh the ball as accurately as a laboratory scale. It is quite enough to weigh with an accuracy of 1 g. It is enough to measure the height of the cone and the radius of the circle with an accuracy of 1 cm. With such measurement accuracy, the relative errors of the quantities will be of the same order.
The order of work.
1. Determine the mass of the ball on the scales with an accuracy of 1 g.
2. We pass the thread through the hole in the cork and clamp the cork in the tripod foot (see Fig. V).
3. Draw a circle on a piece of paper, the radius of which is about 20 cm. We measure the radius with an accuracy of 1 cm.
4. We position the tripod with the pendulum so that the continuation of the thread passes through the center of the circle.
5. Taking the thread with your fingers at the suspension point, rotate the pendulum so that the ball describes the same circle as the one drawn on the paper.
6. We count the time during which the pendulum makes given number revolutions (for example, N = 50).
7. Determine the height of the conical pendulum. To do this, we measure the vertical distance from the center of the ball to the suspension point (we consider h ~ l).
8. Find the module of centripetal acceleration using the formulas:
a n = 4π 2 R/T 2 And a n = gR/h
9. Using a horizontal dynamometer, we pull the ball to a distance equal to the radius of the circle and measure the modulus of the component F 1. Then we calculate the acceleration using the formula and n = F 1 /m.
10. We enter the measurement results into a table.
| Experience no. | R | N | Δt | T = Δt/N | h | m | a n = 4π 2 R/T 2 | a n = gR/h | a n = F 1 /m |
| 1 |
Comparing the obtained three values of the centripetal acceleration module, we are convinced that they are approximately the same.
No. 1. Study of body movement in a circle
Goal of the work
Determine the centripetal acceleration of the ball when it moves uniformly in a circle.
Theoretical part
Experiments are carried out with a conical pendulum. A small ball moves in a circle of radius R. In this case, the thread AB, to which the ball is attached, describes the surface of a right circular cone. From the kinematic relations it follows that аn = ω 2 R = 4π 2 R/T 2.
Two forces act on the ball: the force of gravity m and the tension force of the thread (Fig. L.2, a). According to Newton's second law, m = m +. Having decomposed the force into components 1 and 2, directed radially to the center of the circle and vertically upward, we write Newton’s second law as follows: m = m + 1 + 2. Then we can write: ma n = F 1. Hence a n = F 1 /m.
The modulus of the component F 1 can be determined using the similarity of triangles OAB and F 1 FB: F 1 /R = mg/h (|m| = | 2 |). Hence F 1 = mgR/h and a n = gR/h.
Let's compare all three expressions for a n:
and n = 4 π 2 R/T 2, and n =gR/h, and n = F 1 /m
and make sure that the numerical values of the centripetal acceleration obtained by the three methods are approximately the same.
Equipment
A tripod with a coupling and a foot, a measuring tape, a compass, a laboratory dynamometer, a scale with weights, a ball on a string, a piece of cork with a hole, a sheet of paper, a ruler.
Work order
1. Determine the mass of the ball on a scale with an accuracy of 1 g.
2. Pass the thread through the hole in the plug and clamp the plug in the tripod foot (Fig. L.2, b).
3. Draw a circle on a piece of paper with a radius of about 20 cm. Measure the radius to the nearest 1 cm.
4. Position the tripod with the pendulum so that the continuation of the thread passes through the center of the circle.
5. Taking the thread with your fingers at the suspension point, rotate the pendulum so that the ball describes the same circle as the one drawn on the paper.
6. Count the time during which the pendulum makes a given number (for example, in the range from 30 to 60) revolutions.
7. Determine the height of the conical pendulum. To do this, measure the vertical distance from the center of the ball to the suspension point (we assume h ≈ l).
9. Pull the ball with a horizontal dynamometer to a distance equal to the radius of the circle and measure the modulus of component 1.
Then calculate the acceleration using the formula
Comparing the obtained three values of the centripetal acceleration module, we are convinced that they are approximately the same.
Laboratory work No. 4 in physics, grade 9 (answers) - Study of the motion of a body in a circle
3. Calculate and enter into the table the average value of the time period
4. Calculate and enter into the table the average value of the rotation period
5. Using formula (4), determine and enter into the table the average value of the acceleration module.
6. Using formulas (1) and (2), determine and enter into the table the average value of the angular and linear speed.
| Experience | N | t | T | a | ω | v |
| 1 | 10 | 12.13 | - | - | - | - |
| 2 | 10 | 12.2 | - | - | - | - |
| 3 | 10 | 11.8 | - | - | - | - |
| 4 | 10 | 11.41 | - | - | - | - |
| 5 | 10 | 11.72 | - | - | - | - |
| Wed. | 10 | 11.85 | 1.18 | 4.25 | 0.63 | 0.09 |
7. Calculate maximum value absolute random error in measuring the time interval t.
8. Determine the absolute systematic error of the time period t.
9. Calculate absolute error direct measurement of time interval t.
10. Calculate the relative error of direct measurement of the time interval.
11. Write down the result of direct measurement of a period of time in interval form.
Answer security questions
1. How will the linear speed of the ball change when it rotates uniformly relative to the center of the circle?
Linear speed is characterized by direction and magnitude (modulus). The modulus is a constant quantity, but the direction during such movement can change.
2. How to prove the relation v = ωR?
Since v = 1/T, the relationship between the cyclic frequency and the period is 2π = VT, whence V = 2πR. The connection between linear velocity and angular velocity is 2πR = VT, hence V = 2πr/T. (R - radius of the described, r - radius of the inscribed)
3. How does the rotation period T of the ball depend on the magnitude of its linear velocity?
The higher the speed indicator, the lower the period indicator.
Conclusions: learned to determine the period of rotation, modules, centripetal acceleration, angular and linear speed at uniform rotation body and calculate the absolute and relative errors of direct measurements of the time interval of body movement.
Super task
Determine Acceleration material point with its uniform rotation, if in Δt = 1 s it has covered 1/6 of the circumference, having a linear velocity module v = 10 m/s.
Circumference:
S = 10 ⋅ 1 = 10 m
l = 10⋅ 6 = 60 m
Circle radius:
r = l/2π
r = 6/2 ⋅ 3 = 10 m
Acceleration:
a = v 2 /r
a = 100 2 /10 = 10 m/s 2.
Date__________ FI________________________________________ Class 10_____
Laboratory work No. 1 on the topic:
“STUDYING THE CIRCULAR MOTION OF A BODY UNDER THE INFLUENCE OF ELASTICITY AND GRAVITY FORCES.”
Goal of the work: determination of the centripetal acceleration of a ball during its uniform motion in a circle.
Equipment: tripod with coupling and foot, measuring tape, compass, dynamometer
laboratory, scales with weights, weight on a string, sheet of paper, ruler, cork.
Theoretical part of the work.
Experiments are carried out with a conical pendulum. A small ball moves along a circle of radius R. In this case, the thread AB, to which the ball is attached, describes the surface of a right circular cone. There are two forces acting on the ball: gravity 
and thread tension 
(Fig. a). They create centripetal acceleration 
, directed radially towards the center of the circle. The acceleration modulus can be determined kinematically. It is equal to:
.
To determine the acceleration, it is necessary to measure the radius of the circle and the period of revolution of the ball along the circle.
Centripetal (normal) acceleration can also be determined using the laws of dynamics.
According to Newton's second law 
. Let's break down the power into components 
And 
, directed radially to the center of the circle and vertically upward.
Then Newton's second law will be written as follows:
.
We choose the direction of the coordinate axes as shown in Figure b. In projections onto the O 1 y axis, the equation of motion of the ball will take the form: 0 = F 2 - mg. Hence F 2 = mg: component balances gravity , acting on the ball.
Let's write down Newton's second law in projections onto the O 1 x axis: man = F 1 . From here 
.
The modulus of the component F 1 can be determined in various ways. Firstly, this can be done from the similarity of triangles OAB and FBF 1:
.
From here 
And 
.
Secondly, the modulus of the component F 1 can be directly measured with a dynamometer. To do this, we pull the ball with a horizontally located dynamometer to a distance equal to the radius R of the circle (Fig. c), and determine the reading of the dynamometer. In this case, the elastic force of the spring balances the component .
Let's compare all three expressions for a n:
, , and make sure that they are close to each other.
Progress.
1. Determine the mass of the ball on the scale with an accuracy of 1 g.
2. Secure the ball suspended on a thread in the tripod leg using a piece of cork.
3 . Draw a circle with a radius of 20 cm on a piece of paper (R= 20 cm = ________ m).
4. We position the tripod with the pendulum so that the extension of the cord passes through the center of the circle.
5 . Taking the thread with your fingers at the suspension point, bring the pendulum to rotational movement
above a sheet of paper so that the ball describes the same circle as the one drawn on the paper.
6. We count the time during which the pendulum makes 50 full revolutions (N = 50).
7. Calculate the period of revolution of the pendulum using the formula: T = t / N.
8 . Calculate the value of centripetal acceleration using formula (1):
=
9 . Determine the height of the conical pendulum (h). To do this, measure the vertical distance from the center of the ball to the suspension point.
10 . Calculate the value of centripetal acceleration using formula (2):
=
11.
Pull the ball with a horizontal dynamometer to a distance equal to the radius of the circle and measure the modulus of the component .
Then we calculate the acceleration using formula (3): =
12. The results of measurements and calculations are entered into the table.
| Circle radius R , m | Speed N | t , With | Circulation period T = t / N | Pendulum height h , m | Ball mass m , kg | Center acceleration m/s 2 | Center acceleration m/s 2 | Center acceleration m/s 2 |
13 . Compare the obtained three values of the centripetal acceleration module.
__________________________________________________________________________ CONCLUSION:
______________________________________________________________________________________________________________________________________________________________________________________________________________________________
Additionally:
Find the relative and absolute error indirect measurement a c (1) and (3):
Formula 1). ________ ; Δa c = · a c = ________;
Formula (3). _________; Δa c = · a c = _______.