Lecture 10

Employ Newton’s Laws in 2D problems with circular motion

Uniform Circular Motion

For an object moving along a curved trajectory, with non-uniform speeda = ar (radial only)

ar

v|ar |=

v2T

r

Perspective is important

Non-uniform Circular Motion

For an object moving along a curved trajectory, with non-uniform speeda = ar + at (radial and tangential)

ar

at

dt

d| |v|aT | =

|ar |= v2

T

r

Circular motion

Circular motion implies one thing

|aradial | =v2T / r

Key steps

Identify forces (i.e., a FBD)

Identify axis of rotation

Apply conditions (position, velocity, acceleration)

Example

The pendulum

Consider a person on a swing:

When is the tension on the rope largest? And at that point is it :

(A) greater than(B) the same as(C) less than

the force due to gravity acting on the person?

axis of rotation

Example Gravity, Normal Forces etc.

vT

mg

T

at bottom of swing vT is max

Fr = m ar = m vT2 / r = T - mg

T = mg + m vT2 / r

T > mg

mg

T

at top of swing vT = 0

Fr = m 02 / r = 0 = T – mg cos

T = mg cos T < mg

axis of rotation

y

x

Conical Pendulum (very different) Swinging a ball on a string of length L around your head

(r = L sin ) axis of rotation Fr = mar = T sin

Fz = 0 = T cos – mg

so

T = mg / cos (> mg)

mar = mg sin / cos

ar = g tan vT2/r vT = (gr tan )½

Period:t = 2 r / vT =2 (r cot

/g)½

A match box car is going to do a loop-the-loop of radius r.

What must be its minimum speed vt at the top so that it

can manage the loop successfully ?

Loop-the-loop 1

To navigate the top of the circle its tangential velocity vT must be such that its centripetal acceleration at least equals the force due to gravity. At this point N, the normal force, goes to zero (just touching).

Loop-the-loop 1

Fr = mar = mg = mvT2/r

vT = (gr)1/2

mg

vT

The match box car is going to do a loop-the-loop. If the speed at the bottom is vB, what is the normal force, N, at that point?

Hint: The car is constrained to the track.

Loop-the-loop 2

mg

v

N

Fr = mar = mv2B/r = N - mg

N = mvB2/r + mg

Once again the car is going to execute a loop-the-loop. What must be its minimum speed at the bottom so that it can make the loop successfully?

This is a difficult problem to solve using just forces. We will skip it now and revisit it using energy considerations later on…

Loop-the-loop 3

Example, Circular Motion Forces with Friction (recall mar = m |vT |

2 / r Ff ≤ s N )

How fast can the race car go?

(How fast can it round a corner with this radius of curvature?)

mcar= 1600 kgS = 0.5 for tire/road r = 80 m g = 10 m/s2

r

Only one force is in the horizontal direction: static friction

x-dir: Fr = mar = -m |vT | 2 / r = Fs = -s N (at maximum)

y-dir: ma = 0 = N – mg N = mg

vT = (s m g r / m )1/2

vT = (s g r )1/2 = (x 10 x 80)1/2

vT = 20 m/s

Example

N

mg

Fs

mcar= 1600 kgS = 0.5 for tire/road r = 80 m g = 10 m/s2

y

x

Acceleration, Force, Velocity

F av

Zero Gravity Ride

A rider in a “0 gravity ride” finds herself stuck with her back to the wall.

Which diagram correctly shows the forces acting on her?

Banked Curves

In the previous car scenario, we drew the following free body diagram for a race car going around a curve on a flat track.

n

mg

Ff

What differs on a banked curve?

Banked Curves

Free Body Diagram for a banked curve.

Use rotated x-y coordinates

Resolve into components parallel and perpendicular to bank

N

mgFf

For very small banking angles, one can approximate that Ff is parallel to mar. This is equivalent to the small angle approximation sin = tan , but very effective at pushing the car toward the center of the curve!!

mar

xx y y

Banked Curves, Testing your understanding

Free Body Diagram for a banked curve.

Use rotated x-y coordinates

Resolve into components parallel and perpendicular to bank

N

mgFf

At this moment you press the accelerator and, because of the frictional force (forward) by the tires on the road you accelerate in that direction.

How does the radial acceleration change?

mar

yy

x x