Chapter 5
The Giancoli onLine Tutor

Circular Motion; Gravitation

I. Circular Motion and Centripedal Acceleration

When you see circle or circular in a problem that is a sure sign that you are going to need to use the formula for centripetal acceleration:

a_c=v²/R

v is the speed at that instant

R is the radius of the circle

The centripetal acceleration is always directed towards the center of the circle. Sometimes the term centripetal force is used but this can be misleading to the student as it is not a "new" force but, rather, whatever force or forces give rise to the circular motion.

Some examples:

Hockey puck on string whirled in a horizontal circle on a frictionless surface
Moon going in a "circle" around the earth
Airplane executing a vertical circle
A car going in a circle on a flat track

The "centripetal Forces" for each of these are:

The Tension in the string
Gravity
A vector combination of the
- weight: w
- wing lift:F_L
- engine thrust:F_T
friction between the tires and the road

Play the movie at the left. After watching it for a couple of times, use the right hand buttons to step back and forth through the four scenes. In all the scenes ma_c is a vector pointed in the direction of the acceleration and of magnitude of m*|a_c|. It is of course the right hand side of (Newton's Second Law):
F=ma
The subscript "c" tells you it is a centripetal acceleration with magnitude v²/R. The plane is going faster at the bottom than anyplace else and the mv²/R consequently is longest there.

Film Frame	The left hand side of Newton's Second Law
Bottom	\|F_L\|-\|w\|
Going Up	\|F_L\|
At the Top	\|F_L\|+\|w\|
Coming Down	\|F_L\|

Where F_L is the lifting force and w is the weight or gravitational force and in all cases the right hand side is mv²/R!

Note if you do not have the quicktime plugin you can view a GIF animation version of the movie here.

Tangential Acceleration

In most problems the speed of an object travelling in a circle will be constant and the only acceleration the object will be its centripetal acceleration. In this case the plane also has a tangential acceleration (an acceleration parallel to its motion which increases or decreases its speed). In the forth frame, even without thrust from the planes engines, the plane is being accelerated downward by its weight, which is parallel to the motion at this point:

w=ma_t.

In the second frame the plane is slowing down because the thrust, F_T, up is less then the weight, w, down.

Forces (or components) parallel to the motion produce a tangential acceleration.
Forces (or components) perpendicular to the motion produce a centripetal acceleration.

Interactive Example
A roller coaster car negotiates a vertical loop-the-loop

ANSWER:

Strategies

All the problems here that deal with circular motion, only the centripetal acceleration is important. You can work the problems by identifying the actual force (or forces) which is (are) producing the centripetal acceleration and setting it (them) equal to mv²/R.

The basic variables are:

Interactive Example
A car barely negotiates (without slipping) a curve with a radius of curvature of 100 m at 72 kmh

ANSWER:

Unless you are dealing with Newton's Law of Gravity (more about that later) F=ma is the only equation we have to use, so it is necessary that we know three of the variables to find the forth. Sometimes the Period, T, of the motion (the time that it takes the object to complete one complete circle of its motion) will be given instead of v or R. In that case, you can find the unknown one from the relationship:

v= (2*pi*R)/T (speed equals distance/time)

II. Newton's Law of Universal Gravity

F=	Gm₁m₂

	R²

Most commonly known celestial motions involve almost circular orbits:

Earth around Sun
Moon around Earth
Binary Stars

and can be idealized as circular motion.

Binary Star

In the movie at the left, which represents a binary star system, the red "star" is twice as massive as the blue "star" and they orbit a point called the center of mass (more about that in a later chapter) located 2/3 of the way between the blue star and the red star. In this case the orbits are circular. The red arrow represents the gravitational force the blue star exerts on the red one, and the blue arrow is the force that the red one exerts on the blue star. They are equal and opposite as required by Newton's Third Law. In this situation it is this force which is the centripetal force.

Note: If you do not have the quicktime plugin you can view a GIF animation version of the movie here .

The magnitude of the force is the same on each star, since the one star has a mass that is twice that of the other, it must have one half the acceleration.

F1= m_ra_r F2=m_ba_b F1=F2 =>m_ra_r=m_ba_b m_r=2m_b =>a_r=.5a_b

That is why the red star orbits in a smaller circle!

Any problem you encounter will probably be much less complicated than the Binary star problem, usually involving a single "small" body orbiting a single massive and hence "unmoving" central body.

As an illustration we will work the binary star problem with the following assumptions:

The lighter star has a mass of 1*10³⁰ kg
The distance between them is 1.5*10¹¹ m

The purpose is to find the period, T, of the stars (the time it takes for one complete revolution) motion.

Recall that T=(2*p*R)/v
Since we know the orbital radius of each star:

R_b=.67*1.5*10¹¹ m=1*10¹¹ m
R_r=.33*1.5*10¹¹ m= .5*10¹¹ m

We need only find the velocity of either body to find the period. To do this we can use Newton's Law of Gravity:

F=	Gm_rm_b

	R²

where G is the Universal Gravitational Constant (6.67x10^{-1 1} N*m^2/kg^2).
and R is the distance between the bodies (1.5*10¹¹ m)
Since this is the central force acting on either body it must equal the mass times the centripetal acceleration of either one. For the red body it must equal:

m_r*v²

R²_r

Note: m_r will cancel out when they are set equal and this will always be the case. The mass of the moving body will cancel out. You don't need to know it!
You end up with:
v²=G*m_b*R_r/R²

Note: In most problems the radius of the orbit (R_r) and the distance between the bodies (R) will be the same.

For the numbers we are using here the above formula will yield a velocity of 12000 m/s and from that we can calculate the period ( 2.6x10⁷ sec or about 1 year).

Satellites and Circular Motion

Problems involving the earth and satellites can be simplified by using the equality:

g=	G*m_E

	R_E²

Using this you can rewrite Newton's Second Law as:

F = m_sg (R/R_E)²

Divide R_E by R, square the result and multiply by g*m_s

In most problems F will be The Gravitational Force and m_s will cancel out.

m_sg (R_E/R)² = m_sv_s²/R From which we find.

v_s² = gR(R_E/R)²

Interactive Example 3
A 2000-kg satellite orbits the earth in a a circular orbit 6400 km above the earths surface (earth radius=6400 km).

ANSWER: