Section 7-1 Circular Motion

 

Acceleration can be produced by a change in magnitude of the velocity and/or by a change in ____________________ of the velocity.  Remember, velocity is a vector.  A car moving on a circular path at constant speed has acceleration due to the change in direction.   This type of acceleration is called _______________________ (center-seeking) acceleration, ac.

Centripetal acceleration is the acceleration directed toward the ____________ of a circular path.

ac = vt2 / r  

A car is moving around a circular path.It speeds can be increasing or decreasing.

The car has a component of centripetal acceleration,since the car is continually changing ______________.

The car has a tangential component of acceleration,since the car’s speed is increasing or ___________________.

Ex. 1 A test car moves at a constant speed of 19.7 m/s around a circular track. If the distance from the car to the center of the track is 48.2 m, what is the centripetal acceleration of the car?

Ex. 1 A test car moves at a constant speed of 19.7 m/s around a circular track. If the distance from the car to the center of the track is 48.2 m, what is the centripetal acceleration of the car?

Tangential and centripetal accelerations are ___________________. The total acceleration can be found by using the _______________________ theorem. a2

total = at2 + ac

2 The direction of the total acceleration can be found using trigonometry, tan = ___________

Tangential component is due to the change in speed. Centripetal component is due to change in direction.

Causes of Circular MotionA ball moves in a horizontal circular path with a constant speed.  Thus the magnitude of the velocity is constant, but its direction is constantly changing.  This means that the ball experiences a _________________________acceleration directed toward the center of motion. ac = vt

2 / r

The inertia of the ball tends to maintain the ball's motion in a straight-line path.  But the string counteracts this by exerting a force on the ball that makes the ball follow a _____________ path.

The force is directed along the length of the string toward the center of the circle.  This force can be found by using Newton's ________________ law:  Fc = mac

Fc is called centripetal force and is the force needed to maintain circular motion.  The unit of Fc is the ____________________.

Combining the formulas: Fc = mac ac = vt

2 / r You get: Fc = m(ac) Fc = __________ Fc is the centripetal force that enables a car to travel in a circular path. The ____________ between the car’s tires and the road provides the centripetal force. The ________________ force the earth exerts on the moon provides the centripetal force needed to keep the moon in orbit. This centripetal force acts at ________________ angles to the motion and causes a change in direction of the velocity.

Ex 2 What happens to an object in circular motion when the centripetal force disappears?

Ex 2 What happens to an object in circular motion when the centripetal force disappears?

• The object stops moving in a circular path. The object moves along a straight path that is tangent to the circle.

Ex 3 A 70.5 kg pilot is flying a small plane at 30.0 m/s in a circular path with a radius of 100.0 m. Find the magnitude of the force that maintains the circular motion of the pilot.

Ex 3 A 70.5 kg pilot is flying a small plane at 30.0 m/s in a circular path with a radius of 100.0 m. Find the magnitude of the force that maintains the circular motion of the pilot.

Section 7-2 Newton’s Laws of Universal Gravitation

Gravitational force is the mutual force of ______________________ between particles of matter.  It exists between any two masses regardless of size and composition.

Newton’s law of universal gravitation is stated as the force of attraction between two objects is directly proportional to the product of the _____________ of the objects and inversely proportional to the square of the ___________________ between their centers of mass.

The equation form of the law of universal gravitation is _________________________________ where m1 and m2 are the masses of the attracting objects in kg, r (in m) is the distance between the centers of mass, G is the constant of universal gravitation or gravitation constant, and Fg is the force of gravitation in N. G = 6.673 x 10-11 Nm2/kg2 FEm or force exerted on Earth by the moon FmE or force exerted on Earth by the moon By Newton’s ________________ Law, FmE = FEm.

Earth

Moon

• The universal law of gravitation is an inverse-square law.  This means that the force varies as the inverse square of the separation, so that the force between two masses ____________________ as the masses move father apart.

• F  1/r2

Ex. 4 What is the gravitational force of attraction between two 11.0 kg masses held 3.0 m apart? G = 6.673x 10-11 Nm2/kg2

Ex. 4 What is the gravitational force of attraction between two 11.0 kg masses held 3.0 m apart? G = 6.673x 10-11 Nm2/kg2

Ex. 5 Determine the force that maintains the circular motion of Mercury (3.18 x 1023 kg) around the Sun (1.991 x 1030 kg). Mercury orbits the Sun at a range of 5.79 x 1010 m.

Ex. 5 Determine the force that maintains the circular motion of Mercury (3.18 x 1023 kg) around the Sun (1.991 x 1030 kg). Mercury orbits the Sun at a range of 5.79 x 1010 m.

Section 7-3 Motion in SpaceOrbital Motion for celestial bodies can be circular, elliptical, or parabolic.  For circular orbital motion, centripetal force is provided by gravitational forces. Kepler’s 3 Laws of Motion

Kepler’s 3 Laws of Motion

Kepler’s 3rd Law of Motion:  T2 = 4π2r3 / GM (M is mass of sun.)  

Kepler’s Laws apply to any body that orbits the sun.

Kepler’s 3rd Law is also valid for an elliptical orbit, provided r represents the average distance from the sun.