College Physics I: BIIG problem-solving method

Rotation

• The simplest form of curved motion is uniform circular motion, motion in a circular path at constant speed.
• Pure rotational motion occurs when points in an object move in circular paths centered on one point.
• Pure translational motion is motion with no rotation.
• Some motion combines both types, such as a rotating hockey puck moving along ice.

Rotation Angle

• The rotation angle is the amount of rotation and is analogous to linear distance.
• We define the rotation angle Δθ to be the ratio of the arc length to the radius of curvature:

            Δθ   =   Δs / r

The arc length Δs is the distance traveled along a circular path.

• We know that for one complete revolution, the arc length is the circumference of a circle of radius r.

The circumference of a circle is 2 π r. Thus for one complete revolution the rotation angle is

            Δθ   =   2 π r / r   =   2 π

• The units used to measure rotation angles, Δθ are radians (rad), defined so that

2 π rad   =   1 revolution                  2 π rad  =  360º                  1 rad  =  360º / 2 π  ≈  57.3º

Angular Velocity

• We define angular velocity ω as the rate of change of an angle.

             ω   =   Δθ / Δt

where, an angular rotation Δθ takes place in a time Δt.

The units for angular velocity are radians per second (rad/s).

• Angular velocity ω is analogous to linear velocity v.
• Problem (E6.1):  Calculate the angular velocity of a 0.300 m radius car tire when the car travels at 15.0 m/s (about 54 km/h).                                                                                                              ( 50.0 rad/s )

Centripetal Acceleration

• The magnitude of the centripetal acceleration is

            a_c   =   v² / r

• which is the acceleration of an object in a circle of radius r at a speed v.
• Since, the linear velocity is

            v   =   ω r                     a_c  =  r ω²       

The direction of a_c is toward the center.

• A centrifuge is a rotating device used to separate specimens of different densities.
• Problem (E6.2):  What is the magnitude of the centripetal acceleration of a car following a curve of radius 500 m at a speed of 25.0 m/s (about 90 km/h)?  Compare the acceleration with that due to gravity for this fairly gentle curve taken at highway speed.                                                            ( 0.1 g )
• Problem (E6.3):  Calculate the centripetal acceleration of a point 7.50 cm from the axis of an ultracentrifuge spinning at 7.5 × 10⁴ rev/min. Determine the ratio of this acceleration to that due to gravity.                                                                                                                                        ( 470,000 ) 

Centripetal Force

• Any net force causing uniform circular motion is called a centripetal force.
• The direction of a centripetal force is toward the center of curvature, the same as the direction of centripetal acceleration.
• The magnitude of centripetal force F_c is

            F_c   =   m a_c

Therefore,

            F_c   =   m v² / r      ;        F_c   =   m r ω²

• Problem (E6.4):  Calculate the centripetal force exerted on a 900 kg car that negotiates a 500 m radius curve at 25.0 m/s. Assuming an unbanked curve, find the minimum static coefficient of friction, between the tires and the road, static friction being the reason that keeps the car from slipping.         ( 0.1 ) 

Ideal banking

• For ideal banking, the net external force equals the horizontal centripetal force in the absence of friction.  If the angle θ is ideal for the speed and radius, then the net external force will equal the necessary centripetal force.

            θ   =   tan^-1 ( v² / r g )      (ideally banked curve, no friction)

Note that θ does not depend on the mass of the vehicle.

• Problem (E6.5):  Curves on some test tracks and race courses, such as the Daytona International Speedway in Florida, are very steeply banked. This allows the curves to be taken at very high speed. Calculate the speed at which a 100. m radius curve banked at 65.0° should be driven if the road is frictionless.                                                                                                                                 ( 45.8 m/s ) 

Fictitious Forces

• For example, you make a tight curve in your car to the right. The force to the left sensed by car passengers is a fictitious force having no physical origin.  There is nothing real pushing them left—the car, as well as the driver, is actually accelerating to the right.

Newton’s Universal Law of Gravitation

• Newton’s universal law of gravitation states that every particle in the universe attracts every other particle with a force along a line joining them. The force is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
• For two bodies having masses m and M with a distance r between their centers of mass, the equation for Newton’s universal law of gravitation is

            F   =   G m M / r²

where, F is the magnitude of the gravitational force and G is a proportionality factor called the gravitational constant.

• G is a universal gravitational constant. It is thought to be the same everywhere in the universe.
• In SI units, it has been measured experimentally to be

            G  =  6.673×10^-11  N m² / kg²

Note that the units of G are such that a force is in newtons.

Acceleration due to Gravity

• The weight of an object

            w   =   m g

is the gravitational force between it and Earth.

• Substituting in Newton’s universal law of gravitation gives

            g   =   G M / r²

where, m is the mass of the object, M is the mass of Earth, and r is the distance to the center of Earth.

• Using Earth’s mass and radius

            M  =   5.98×10²⁴  kg                r   =   6.38×10⁶  m

we obtain a value for the acceleration of a falling body

            g   =   9.80 m/s²

• Problem (E6.6):  Earth’s mass is 5.98×10²⁴ kg.  And the radius of the Moon’s nearly circular orbit is 3.84×10⁸ m. Find the acceleration due to Earth’s gravity at the distance of the Moon.            ( 0.00271 m/s² ) 

Weightlessness

• Weightlessness doesn’t mean that an astronaut is not being acted upon by the gravitational force.
• There is no “zero gravity” in an astronaut’s orbit.

The term just means that the astronaut is in free-fall, accelerating with the acceleration due to gravity.

• Microgravity refers to an environment in which the apparent net acceleration of a body is small compared with that produced by Earth at its surface.

The Cavendish Experiment

• Henry Cavendish (1731–1810), an English scientist, in 1798, was the first to accurately determine experimentally the universal gravitational constant G. More than 100 years after Newton published his universal law of gravitation.
• Using

            g   =   G M / r²

Solving for the mass

            M   =   r² g / G

Therefore, knowing G allows for the determination of astronomical masses.

Kepler’s Laws of Planetary Motion

• Kepler’s First Law:  The orbit of each planet about the Sun is an ellipse with the Sun at one focus.
• Kepler’s Second Law:  Each planet moves so that an imaginary line drawn from the Sun to the planet sweeps out equal areas in equal times.
• Kepler’s Third Law:  The ratio of the squares of the periods of any two planets about the Sun is equal to the ratio of the cubes of their average distances from the Sun.

T₁² / T₂² =   r₁³ / r₂³

where, T is the period (time for one orbit) and r is the average radius.

This equation is valid only for comparing two small masses orbiting the same large one.

• Problem (E6.7):  Given that the Moon orbits Earth each 27.3 d and that it is an average distance of 3.84×10⁸ m from the center of Earth, calculate the period of an artificial satellite orbiting at an average altitude of 1500 km above Earth’s surface (6380 km).                                                         ( 1.93 h ) 

BIIG: Problems & Solutions