A 0.5 Kg Object Moves on a Horizontal Circular Track
Force and Acceleration in Circular Motion
Introduction
Acceleration is the time rate of change of velocity. Since velocity is a vector, it can change in two ways: its magnitude can change and its direction can change. Either change gives rise to an acceleration. For circular motion at constant speed, the velocity is always tangential to the circular path, and therefore its direction is continuously changing even though its magnitude is constant. Therefore, the object has an acceleration. It can be shown that the magnitude of the acceleration a c for uniform circular motion with speed v in a path of radius R is a c = ,
Newton's second law requires that there be a net force on the object equal in magnitude to ma c and in the direction of a c. Circular motion with speed v in a path of radius R has period (time for one revolution) T and frequency (revolutions/s) f = 1/ T . v = = 2 π Rf a c = 4 π 2 f 2 R . Σ F = m a F string = Mg ,
Because of the downward force of gravity on the ball, when the ball moves in a horizontal circle the string is at an angle θ below the horizontal, as shown in Figure 3. In the figure, L is the length of the string, measured from the center of the tube to the center of the ball. The radius R of the circular path of the ball is given by R = L cos θ .
The forces on the ball are gravity and the tension in the string. The tension in the string is directed along the string and the gravity force is straight downward. The free-body diagram for the moving ball is given in Figure 4. Since the ball moves in a horizontal circle, its acceleration is horizontal. It is convenient therefore to use coordinates that are horizontal and vertical, and in the force diagram F string has been resolved into its horizontal and vertical components.
Σ F x = ma x F string cos θ = ma c = m 4 π 2 f 2 R . R = L cos θ F string = m 4 π 2 f 2 L . F string = Mg , Mg = m 4 π 2 f 2 L .
f 2 =
Objective
In this experiment we will test the expression for the acceleration of an object moving in uniform circular motion.
Apparatus
Discussion
As discussed in the Introduction for this experiment, application of Newton's second law in the experimental setup yields the following.
( 1 )
f 2 =
In this equation, f is the frequency of the circular motion of the ball and m is the mass of the ball. M is the mass suspended from the lower end of the string and L is the length of string between the center of the ball and the center of the upper end of the tube. Note that the angle θ that the string makes with the horizontal does not appear in Eq. (1). Derivation of Eq. (1) assumes that the acceleration of an object moving at constant speed v in a circular path of radius R has magnitude a c = Σ F = m a.
Procedure
Please print the worksheet for this lab. You will need this sheet to record your data.
1
For practice, place a slotted mass of 100 grams on the mass hanger at the lower end of the string and whirl the racketball over your head while holding onto the string below the tube. Practice spinning the ball over your head while maintaining the path of the ball completely horizontal, until you can let go of the string below the tube and maintain the same motion while the mass does not rise or fall. Each lab partner should do this exercise.
2
Pull enough string through the tube so the length L is 50.0 cm. Recall that L is the distance from the center of the top of the tube to the center of the ball. Attach an alligator clip to the string about 1 cm below the plastic tube to serve as a marker so you can keep L constant while whirling the ball. As you whirl the ball in a horizontal circle, make sure that the string or alligator clip does not come in contact with your hand or arm. Ensure that the ball is rotating in a horizontal circle before measuring the time. Then have your lab partner measure the time t 20(1) it takes for the ball to complete 20 revolutions. Interchange roles of whirler and timer and repeat the measurement to obtain time t 20(2). Repeat taking measurements until you get a pair of times that differ by less than 2.0 seconds.
3
Enter your values of t 20(1) and t 20(2) into the column for M = 150 grams f 1 = , f 2 = , f = .
4
Open Excel and plot f 2 (for f in rev/s) versus in kg/m. Use Excel to find the equation for the straight line that is the best fit to your data. All your data points should fall close to this line and the y-intercept of the line should be close to zero. If one, or both, of these isn't the case, you have made a mistake in collecting, recording or graphing your data. If you and your lab partner can't find what is wrong, ask your TA for help.
5
Record the slope of the line that is the best fit to your data.
6
Use the balance to measure the mass of the ball. Record your results.
7
Use Eq. (1) f 2 =
8
If the actual value of g is taken to be 9.80 m/s2, what is the percentage difference between your experimental result and the actual value of g? In the calculation retain enough significant figures to avoid round-off error.
A 0.5 Kg Object Moves on a Horizontal Circular Track
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