Question

1. Tie one end of the string [~1.5 m] to

one of the small plastic bags into

which you have put 6 US pennies.

We will call the bag/pennies

combination the rotating object, RO.

[A suggested method for making the

RO is given at the end of these inst

ructions.] Pass the other end of the

string through one end of the plastic t

ube. The end of the tube into which

you fed this string is now referred to

as the top end of the tube. Lay the

RO-string-tube combination on a handy fl

at surface. Mo

ve the tube until

its top end is 1.0-m from the center of

the bag of pennies t

hat forms the RO.

[That white-looking stick lying on the

table is a meterstick.] Next, you need to

mark the string at the point where it exits

the bottom of the tube. One way is to

place a paperclip at that point.

Another is to put a drop of paint, nail

polish or liquid paper

[if the string is

dark in color] at that point. Any way you can mark that point so you can

see it as you rotate the Rotating

Object around your head will be fine.

Record the distance from the Rotating Ob

ject to the top of

the tube in the

Radius column for Trial 1. Try to keep

this at 1.0 m for all trials. If for

some reason that is impossible, be sure to record the correct value in the

Radius column.

2. Place the required num

ber of pennies in the second plastic bag and tie the

bag to the string using a slip knot so you can remove the bag easily for the

second trial to add more pennies.

3. Start with 16 pennies as your first ma

ss. The mass, m, is the mass of all

the pennies you have in the plastic bag plus the mass of the paperclip.

Record this mass in kilograms in the

Mass of Pennies column for Trial 1.

[1g=0.001kg] It is the we

ight of these pennies t

hat is transmitted as

tension in the string to keep the rotati

ng object from flying off as you rotate

it. This weight [W=mg] we call the balancing force FB

. Calculate the

weight of the pennies you are using and

record it in the Balancing Force

column.

4. Begin to swing the Rotating Object

around your head. As you swing it

faster the tension in the

string will increase. You must rotate the Rotating

Object at the proper speed so that t

he tension produced by the weight of

the pennies is enough to keep the Rotati

ng Object rotating in a circle of

1.0-m radius. That takes practice!

The more pennies you use the easier it

gets to produce this balanced conditi

on. You’ll use more pennies in

subsequent trials. Make sure that the paperclip you may have used to

mark the bottom end of the tube does not

actually touch the tube. This is

why it is easier to use a spot of whit

e-out or fingernail polish if the string is

light colored. Once you can do this

you’re ready to determine the period

of rotation.

5. Once you have established a balanced co

ndition with the Rotating Object

moving at a constant rate, have your

assistant measure the time recorded

for 20 revolutions of the Rotating Object. Determine the period of a single

rotation by dividing the measured time by the number of revolutions:

20

TimeT=

. Write this in the Period column for trial 1.

6. Now, compute the value

of the centripetal force FC

. In the following

equation the M is the mass

of the Rotating Object

(RO), previously known

as Rotating Object in kilograms.

This mass M will remain constant

throughout the experiment.

7. Calculate the percent error between FC and FB

using the following

equation.()%100 CBB FFerrorF?=×

. If this error is fairly small say 10 % or

less proceed to the next

step. If it is large, 20% or higher you may be

doing something wrong.

8. For Trial 2 add 6 pennies and repeat st

eps 3 through 7. Continue to add 6

pennies for each subsequent trial until

you have completed the 5 trials.

9. Answer the questions and upload

your report to

the Drop Box.

A Discussion of Centripetal Force

According to Newton’s First Law of Mo

tion an object in straight-line motion

at constant speed will continue moving in a

straight-line at constant speed unless

acted on by an external, unbalanced force.

If an external unbalanced force acts

the object will either speed

up, slow down, change direction or any combination

of the first two and third options. If

the force exerted on the object is so

constructed that it can keep a constant

magnitude and stay at right angles to the

direction of the object’s velocity, the

object will move at constant speed in a

circle: uniform circular motion.

Such a force is called a centripetal force because it always points toward

the center of the circle in which

the object is moving. That force FC

is an unbalanced or net force so it satisfies Newton’s Second Law of Motion: F

net =ma.

In this particular case the acceleration

is called the centripetal acceleration.

Centripetal acceleration can be computed from the equation

2cVar=.

Where V is the linear speed of the object and r

is the radius of the circular path.

The amount of time for an object to make

one complete revolution in its circular

path is called the period, T.

The distance around a circle is its circumference

which is given by C= 2?r. The speed is just distance/time so2 circumference

rVtimeT?==

. Substituting this into the equation for centripetal

acceleration gives us22222()4crVrTarrT??== =. Finally multiplying the

acceleration by the mass of the rotating

object [stopper], which we here label as

M, we get224crMFT?=.

Centripetal Force Report

Centripetal Force Experiment Data
Trial	Mass of Pennies (m)	Balancing Force (FB)	Radius (r)	Period (T)	Centripetal Force (Fc)	% error
	(kg)	(N)	(m)	(s)	(N)
1	0.040	0.392	1.0	1.1925	0.4164	6.22
2	0.055	0.539	1.0	1.043	0.5444	1.0
3	0.070	0.686	1.0	0.925	0.6921	0.889
4	0.085	0.833	1.0	0.832	0.8555	2.70
5	0.10	0.980	1.0	0.7605	1.024	4.49
Mass of Rotating Object(M) à		0.015 (kg)

The balancing force is computed using F_B=mg, where m is in kilograms and g is the gravitational acceleration of 9.8 m/s/s.    

The centripetal force is computed using F_C=4p²rM/T²

{Note the different ‘m’ and ‘M’ used here are the two different masses listed in the table. m=mass of pennies in a bag on the lower, vertical end of the string [these produce the balancing force] and M=mass of rotating object used to compute the centripetal force.[should be 6 pennies in a bag]}

Hints: Try to keep the radius the same from trial to trial and about 1 meter. Recall that 1 gram = 0.001 kg. The mass of a post 1983 penny is 2.5 g.

Questions:

1.         In each trial you computed a balancing force and a centripetal force. Were those values, for a particular trial, nearly equal? Should they be? Why or why not?

2.         What experimental factors caused the errors encountered? Be specific and tell which, in your opinion, was the largest source of error. Tell why you chose this source as the largest.

3.         Did the percent errors tend to grow or get smaller as more mass was added to the string? If you see a trend here explain why it should exist.

             4.        Summarize what you’ve learned from this experiment.

Answer 1

1. They are nearly equal but never could be exactly equal. That's because the centripetal force is never equal to the tension on the string, the centripetal force is a horizontal component of the tension. The string never makes a right angle while rotating, so does the tension vector along it. That's why the horizontal component (the centripetal force) never becomes equal to it .

2. The radius is the largest source of error. Thats because when the mass rotates the center of revolution never stays still, it wobbles and makes the effective r iduced with error.

3. As more and more mass gets added to the bag the easier it gets to produce this balanced condition. Thats why the wobble and the error associated with it should decrease. For the first 3 observations we can see these trend, (last two are caused by random errors )

4. We've learned that centrepetal force is a special state of newton's 2nd law of motion. It's caused by a rotating object.