Rotational Motion and the Law of Gravity Lecture Notes Physics 2053 Rotational Motion and the Law of...

Rotational Motion andthe Law of Gravity

Lecture NotesPhysics 2053

Rotational Motion and the Law of Gravity


7-04 Centripetal Acceleration

Topics

7-05 Newtonian Gravitation

7-06 Kepler’s Laws


Centripetal Acceleration

Uniform circular motion: motion in a circle of constant radius at constant speed

v1

v2

Instantaneous velocity is always tangent to circle.


r

v

rv

rΔvΔ

rrΔv

vΔ

tΔrΔ

rv

tΔvΔ

rv

a2

r

Radial Acceleration:

r2

r1

v1

v2

v2v1

t

Similar Triangles

Divide by time

CentripetalAcceleration



In uniform circular motion the acceleration is called the centripetal, or radial, acceleration. It is perpendicular to the velocity and points towards the center of the circle.

v

ar rv

a2

r

r



Is it possible for an object moving with a constant speed to accelerate? Explain.

A) Yes, although the speed is constant, the direction of the velocity can be changing.

B) No, if the speed is constant then the acceleration is equal to zero.

C) No, an object can accelerate only if there is a net force acting on it.

D) Yes, if an object is moving it can experience acceleration

Rotational Motion and The Law of Gravity


A jet plane travelling 525 m/s pulls out of a dive by moving in an arc of radius 6.00 km. What is the plane’s acceleration?

Centripetal Acceleration Problem


An object moves in a circular path at a constant speed. Compare the direction of the object's velocity and acceleration vectors.

A) The vectors are perpendicular.

B) Both vectors point in the same direction.

C) The vectors point in opposite directions.

D) The question is meaningless, since the acceleration is zero.



What type of acceleration does an object moving with constant speed in a circular path experience?

A) free fall

B) constant acceleration

C) linear acceleration

D) centripetal acceleration




For an object to be in uniform circular motion, there must be a net force acting on it.

v

ar

Fr

rv

mmaF2

rrThe radial force

on the ball is provided by the string

There is no centrifugal force acting on the ball

This radial force is called a

centripetal force

rv

mF2

r


The speed of an object inUniform Circular Motion

maF

rmaT

rv

a2

r

r

MgT rmv

Mg2

mMgr

v

Mg

Tm

M



A 0.45 kg ball, attached to the end of a horizontal cord, is rotated in a circle of radius 1.3 m on a frictionless horizontal surface. If the cord will break when the tension in it exceeds 75 N, what is the maximum speed the ball can have?



r

T mg

Motion in a vertical circle

The tension in the string when the ball is at the top.

maF

rv

mTmg2

top

mgr

mvT

2

top

rv

a2

r



r

T

mg

Motion in a vertical circle

maF

rv

mmg T2

bottom

mg r

mvT

2

bottom

The tension in the string when the ball is at the bottom.

rv

a2

r



A bucket of mass 2.00 kg is whirled in a vertical circle of radius 1.10 m. At the lowest point of its motion the tension in the rope supporting the bucket is 25.0 N. (a) Find the speed of the bucket.



A bucket of mass 2.00 kg is whirled in a vertical circle of radius 1.10 m. (b) How fast must the bucket move at the top of the circle so that the rope does not go slack?

Centripetal Acceleration Problem (con’t)


A pilot executes a vertical dive then follows a semi-circular arc until it is going straight up. Just as the plane is at its lowest point, the force on him is

A) less than mg, and pointing up.

B) less than mg, and pointing down.

C) more than mg, and pointing up.

D) more than mg, and pointing down.



ax

0Fy 0mgN

mgN xx maF

r

mvf

2max

max

Nfmax

mgfmax

r

mvmg

2max

grvmax

r

va

2max

x

r

Maximum Speed in horizontal turn



N

mg

fmax

A car goes around a curve of radius r at a constant speed v. What is the direction of the net force on the car?

A) toward the curve's center

B) away from the curve's center

C) toward the front of the car

D) toward the back of the car



A car goes around a curve of radius r at a constant speed v. Then it goes around a curve of radius 2r at speed 2v. What is the centripetal acceleration of the car as it goes around

the second curve, compared to the first?

A) four times as big

B) twice as big

C) one-half as big

D) one-fourth as big



How large must the coefficient of static friction be between the tires and the road if a car is to round a level curve of radius 85 m at a speed of 95 km/h?



a0Fy sinfmgcosN

N

cos

sinfmgN

maFx cosfsinN

rmv

cosfsincos

sinfmg 2

fsinmgcosr

vm

2

sinmgcos

rmv

f 2

r

v

f

Friction on a banked road

mgr

mv2



a

N

mg

sinmgcosr

mvf

2

v

Friction on a banked road

When

sinmgcosr

mv2

f

When

sinmgcosr

mv2

No Friction

When

sinmgcosr

mv2

f



a

N

mg

sinmgcosr

mvf

2

sinmgcosr

mv0

2

sinmgcosr

mv2

rgv

tan2

rgv

tan 2

1

vTurning a banked curve with no friction



The banking angle in a turn on the Olympic bobsled track is not constant, but increases upward from the horizontal.

Coming around a turn, the bobsled team will intentionally "climb the wall," then go lower coming out of the turn. Why do they do this?

A) to give the team better control, because they are able to see ahead of the turn

B) to prevent the bobsled from turning over

C) to take the turn at a faster speed

D) to reduce the g-force on them



Weight

mg

N

Scale

Reading on scale is the normal force

0Fy

0mgN

mgN



mg

NN

At the North Pole:

0F

0mgNN

mgNN

NE

mg

Apparent Weight at the Earth’s Surface

At the Equator:

rmaF

Rmv

Nmg2

E

Rmv

mgN2

E

v



A space station is in the shape of a hollow ring 450 m in diameter. Gravity is simulated by rotating the ring.Find the speed in revolutions per minute needed inorder to simulate the Earth’s gravity.

v

R

N



v

rmaF

NRmv

mgN2

gRv

TR2

v Rf2

gRRf2

R4

gf

2

2254

8.92

rev/s 00332.0

mins 60

srev

00332.0minrev

99.1

rv

a2

r

The speed in revolutions per minute


R = 225 m


Gravitational Force:

Newtonian Gravitation

Gravitational Force is the mutual force of attraction between any two objects in the Universe.

R

m

M

F

F2R

MmGF

Universal Gravitational Constant

2

211

kg

Nm 10 x 67.6G



Gravitational Potential Energy associated with an object of mass m at a distance r from the center of the Earth is

[7.21] r

mmGPE E

r

m

ME



Escape velocity

E

Eesc R

GM2v

escv

ER EMAn object projected upward from the Earth’s surface with a large enough speed will soar off into space and never return.This speed is called the Earth’s escape velocity.

0PEKE ii

0R

mGM2

mv

E

E2esc

mi/h 25,000

km/s 2.11


Calculate the acceleration due to gravity on the Moon. The

Moon’s radius is 1.74 x 106 m and its mass is 7.35 x 1022 kg.

Newtonian Gravitation Problem


A hypothetical planet has a mass 1.66 times that of Earth, but the same radius. What is g near its surface?



Speed of a Satellite

R

m

M

F

F

2R

MmGF v

Rv

mF2

Rv

mR

MmG

2

2

RGM

v



Two planets have the same surface gravity, but planet B has twice the radius of planet A. If planet A has mass m, what is the mass of planet B?

4m D) m 2 C)

m B) 2m

)A



A certain neutron star has five times the mass of our Sun packed into a sphere about 10 km in radius. Estimate the surface gravity on this monster.




The satellite is kept in orbit by its speed – it is continually falling, but the Earth curves from underneath it.

Earth

Withoutgravity

Withgravity


Compared to its mass on the Earth, the mass of an object on the Moon is

A) the same.

B) less.

C) more.

D) half as much.



Calculate the force of Earth’s gravity on a spacecraft 12,800 km (2 Earth radii) above the Earth’s surface if its mass is 1350 kg.



Kepler’s Laws

Kepler’s Three Laws:

1. All planets move in elliptical orbits with the Sun at one of the focal points.


Kepler’s Laws

Area 1 Area 2

Area 1 = Area 2

2. A line drawn from the Sun to any planet sweeps out equal areas in equal time intervals.



Kepler’s Laws


3. The square of the orbital period of any planet is proportional to the cube of the average distance from the planet to the Sun.

T

R

constantR

T3

2


Kepler’s Laws

T

rSM F

maF

rv

mr

mMG

2

2S

2

2S

rT

r2

mr

mMG

2

32

ST

r4GM

2S

2

3

4

GM

T

r


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