Chapter 5 Circular Motion; Gravitation Objectives Identify the force that is the cause of the centripetal acceleration and determine the direction of

Chapter 5

Circular Motion; Gravitation

Objectives

• Identify the force that is the cause of the centripetal acceleration and determine the direction of the acceleration vector.

• Use Newton's laws of motion and the concept of centripetal acceleration to solve word problems.

5-3 Highway Curves, Banked and Unbanked

When a car goes around a curve, there must be a net force towards the center of the circle of which the curve is an arc. If the road is flat, that force is supplied by friction.


If the frictional force is insufficient, the car will tend to move more nearly in a straight line, as the skid marks show.


As long as the tires do not slip, the friction is static. If the tires do start to slip, the friction is kinetic, which is bad in two ways:

1. The kinetic frictional force is smaller than the static.

2. The static frictional force can point towards the center of the circle, but the kinetic frictional force opposes the direction of motion, making it very difficult to regain control of the car and continue around the curve.

Car Negotiating a Flat Turn

R

v

What is the direction of the What is the direction of the force force ON the car?the car?

Ans.Ans. Toward CenterToward Center

FFcc

This central force is exerted This central force is exerted BY BY the road the road ONON the car. the car.

Car Negotiating a Flat Turn

R

v

Is there also an Is there also an outward outward force force acting acting ON ON the car?the car?

Ans.Ans. No,No, but the car does exert a but the car does exert a outward outward reactionreaction force force ON ON the road. the road.

FFcc

Car Negotiating a Flat TurnCar Negotiating a Flat Turn

The centripetal force The centripetal force FFcc is that of static is that of static friction ffriction fss::

The central force The central force FFCC and the friction force and the friction force ffss are not two different forces that are are not two different forces that are equal. There is just equal. There is just oneone force on the car. force on the car. The The naturenature of this central force is static of this central force is static friction.friction.

The central force The central force FFCC and the friction force and the friction force ffss are not two different forces that are are not two different forces that are equal. There is just equal. There is just oneone force on the car. force on the car. The The naturenature of this central force is static of this central force is static friction.friction.

Fc = fsR

v

mFc

n

mg

fs

R

Finding the maximum speed for negotiating a turn without slipping.

Finding the maximum speed for negotiating a turn without slipping.

Fc = fsfs = smgFc =

mv2

R

The car is on the verge of slipping when The car is on the verge of slipping when FFC C is equal to the maximum force of is equal to the maximum force of

static friction static friction ffss. .

R

vm

Fc

Fc = fsn

mg

fs

R

Maximum speed without slipping (Cont.)Maximum speed without slipping (Cont.)

Fc = fs

mv2

R= smg

v = sgR

Velocity v is maximum speed for no slipping.

Velocity v is maximum speed for no slipping.

n

mg

fs R

R

vm

Fc

Example:Example: A car negotiates a turn of A car negotiates a turn of radius radius 70 m70 m when the coefficient of when the coefficient of static friction is static friction is 0.70.7. What is the . What is the maximum speed to avoid slipping?maximum speed to avoid slipping?

v = 21.9 m/s

v = 21.9 m/s(0.7)(9.8)(70m)sv gR

R

v

m Fc

s = 0.7

fs = smgFc = mv2

R

From which:v = sgR

g = 9.8 m/s2; R = 70 m

April 21, 2023

Level Curves – Try this one.• A 1500 kg car moving on a

flat, horizontal road negotiates a curve as shown. If the radius of the curve is 35.0 m and the coefficient of static friction between the tires and dry pavement is 0.523, find the maximum speed the car can have and still make the turn successfully.rgv

April 21, 2023

Level Curves• The force of static friction directed toward

the center of the curve keeps the car moving in a circular path.

rgv smmsm

grm

mgr

m

Nrv

mgN

mgNFr

vmNf

sss

y

ss

/4.13)0.35)(/8.9)(523.0(

0

2

max

2max

max,

Period 1 stopped here

slow speedslow speed

Optimum Banking AngleOptimum Banking AngleBy banking a curve at By banking a curve at thethe optimum angle, the optimum angle, the

normal force normal force nn can can provide the necessary provide the necessary centripetal force without centripetal force without the need for a friction the need for a friction force.force.

fast speedfast speed optimumoptimum

nfs = 0

w w

nfs

w

nfs

R

v

mFc

Free-body DiagramFree-body Diagram

nn

mgmg

Acceleration Acceleration aa is toward is toward the center. Set the center. Set xx axis axis along the direction of along the direction of aacc , i. e., horizontal (left , i. e., horizontal (left to right).to right).

nn

mgmg

nn sin sin

nn cos cos

++ aacc

n

mg

xx

Optimum Banking Angle Optimum Banking Angle (Cont.)(Cont.)

n

mgmg

nn sin sin

nn cos cos

FFxx = m = maacc

FFyy = 0 = 0 n n cos cos = mg= mg

mvmv22

RRnn sin sin Apply Apply

NewtonNewton’’s s 2nd Law to x 2nd Law to x and y axes.and y axes.

n

mg


n cos = mg

mv2

Rn sin

n

mg

2

2

tan

1

mvvR

mg gR

n

mg

n sin

n cos

sintan

cos

nn


n

mg

n sin

n cos

n

mg

2

tanv

gR

Optimum Banking Angle

Example 5:Example 5: A car negotiates a turn A car negotiates a turn of radius 80 m. What is the of radius 80 m. What is the optimum banking angle for this optimum banking angle for this curve if the speed is to be equal to curve if the speed is to be equal to 12 m/s?12 m/s?

n

mg

n sin

n cos

tan = =v2

gR

(12 m/s)2

(9.8 m/s2)(80 m)

tan = 0.184

n

mg

2

C

mvF

R

2

C

mvF

R

How might you find the centripetal force on the car, knowing its mass?

= 10.40


Banking the curve can help keep cars from skidding. In fact, for every banked curve, there is one speed where the entire centripetal force is supplied by the

horizontal component of the normal force, and no friction is required. This occurs when:

April 21, 2023

Banked Curves• A car moving at the

designated speed can negotiate the curve. Such a ramp is usually banked, which means that the roadway is tilted toward the inside of the curve. Suppose the designated speed for the ramp is to be 13.4 m/s and the radius of the curve is 35.0 m. At what angle should the curve be banked?

April 21, 2023

Banked Curves

6.27))m/s 8.9)(m 0.35(

m/s 4.13(tan

tan

cos

0cos

sin

m 0.35 m/s 4.13

21

2

2

rg

v

mgn

mgnFr

mvmanF

rv

y

cr

Show work for problem

frictionless

mg

N

Fy 0 N cos mg

Fr ma N sin mv2 / R

tan v2gR

Rv2g tan

A car of mass, m, is traveling at a constant speed, v, along a curve that is now banked and has a radius, R. What bank angle, , makes reliance on friction unnecessary?

homework

• Questions p.129 # 8, 9

• Problems 9, 10, and do example 5-7 on page 114 (cover up answer)

5-4 Nonuniform Circular Motion

If an object is moving in a circular path but at varying speeds, it must have a tangential component to its acceleration as well as the radial one.

5-4 Nonuniform Circular Motion

This concept can be used for an object moving along any curved path, as a small segment of the path will be approximately circular.

Tangential & Total Acceleration

Tangential & Total Acceleration An object may be

changing its speed (speeding up or slowing down) as it moves in a circular path. In that case, there is a tangential acceleration as well as a centripetal acceleration.

The total acceleration atotal is the vector sum of the centripetal acceleration acp, which points toward the center of rotation, and the tangential acceleration at, which points in the direction of speed increase.

Tangential Acceleration

• Tangential Acceleration – The instantaneous linear acceleration of an object directed along the tangent to the object’s circular path.

• atan = Δv/Δt

Example 5-8

Example 5-8 Solution

The CentrifugeThe Centrifuge

The apparent weight of an object can be greatly increased using circular motion. A centrifuge is a laboratory device used in chemistry, biology, and medicine for increasing the sedimentation rate and separation of a sample by subjecting it to a very high centripetal acceleration. Accelerations on the order of 10,000 g can be achieved. This can give a 12 g sample can be given an apparent weight of about 1130 N = 250 lb.

5-5 Centrifugation

A centrifuge works by spinning very fast. This means there must be a very large centripetal force. The object at A would go in a straight line but for this force; as it is, it winds up at B.

Example: Big Gees

A centrifuge rotates at a rate such that the bottom of a test tube travels at a speed of 89.3 m/s. The bottom of the test tube is 8.50 cm from the axis of rotation. What is the centripetal acceleration acp at the bottom of the test tube in m/s and in g (where 1 g = 9.81 m/s2)?

2 22(89.3 m/s)

93,800 m/s 9,560 g(0.0850 m)cp

va

r

“Centrifugal Force”

• “Centrifugal force” is a fictitious force - it is not an interaction between 2 objects,

and therefore not a real force.

• Nothing pulls an object away from the center of the circle.

“Centrifugal Force”

• What is erroneously attributed to “centrifugal force” is actually the action of the object’s inertia - whatever velocity it has (speed + direction) it wants to keep.

Documents

Chapter 5 Circular Motion; Gravitation Objectives Identify the force that is the cause of the centripetal acceleration and determine the direction of