physics 207: lecture 10, pg 1 lecture 10 l goals: exploit newton’s 3 rd law in problems with...

Physics 207: Lecture 10, Pg 1

Lecture 10

Goals:Goals:

Exploit Newton’s 3rd Law in problems with friction

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

Assignment: HW5, (Chapter 7, due 2/24, Wednesday)

For Tuesday: Finish reading Chapter 8, First four sections in Chapter 9


Example: Friction and Motion

A box of mass m1 = 1 kg is being pulled by a horizontal string having tension T = 40 N. It slides with friction

(k= 0.5) on top of a second box having mass m2 = 2 kg, which in turn slides on a smooth (frictionless) surface. What is the acceleration of the bottom box ?

Key Question: What is the force on mass 2 from mass 1?

mm22

T mm11slides with friction (k=0.5)

slides without frictiona = ?

v


ExampleSolution

First draw FBD of the top box:

m1

N1

m1g

T fk = KN1 = Km1g

v


Newtons 3rd law says the force box 2 exerts on box 1 is equal and opposite to the force box 1 exerts on box 2.

m1ff1,2 = Km1g = 5 N

m2

f2,1 = -f1,2

As we just saw, this force is due to friction:

ExampleSolution

ActionReaction


Now consider the FBD of box 2:

m2 f2,1 = km1g

m2g

N2

N1 = m1g

ExampleSolution


Finally, solve Fx = max in the horizontal direction:

m2 f2,1 = Km1g

K m1g = m2ax

ExampleSolution

= 2.5 m/s2

kg 2N 5g

a2

k1

mm

x


Home Exercise Friction and Motion, Replay

A box of mass m1 = 1 kg, initially at rest, is now pulled by a horizontal string having tension T = 10 N. This box (1) is on top of a second box of mass m2 = 2 kg. The static and kinetic coefficients of friction between the 2 boxes are s=1.5 and k= 0.5. The second box can slide freely (frictionless) on an smooth surface.

Question:

Compare the acceleration of box 1 to the acceleration of box 2?

m2

T m1

friction coefficients s=1.5 and k=0.5

slides without frictiona2

a1


Exercise Tension example

A. Ta = ½ Tb

B. Ta = 2 Tb

C. Ta = Tb

D. Correct answer is not given

Compare the strings below in settings (a) and (b) and their tensions.


Problem 7.34 Hint (HW 6)

Suggested Steps Two independent free body diagrams are necessary Draw in the forces on the top and bottom blocks Top Block

Forces: 1. normal to bottom block 2. weight 3. rope tension and 4. friction with bottom block (model with sliding)

Bottom Block Forces:

1. normal to bottom surface 2. normal to top block interface3. rope tension (to the left)4. weight (2 kg)

5. friction with top block 6. friction with surface7. 20 N

Use Newton's 3rd Law to deal with the force pairs (horizontal & vertical) between the top and bottom block.


On to Chapter8 Reprisal of : Uniform Circular Motion

For an object moving along a curved trajectory with constant speeda = ar (radial only)

|ar |= vt

2

r

ar

v


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 |= vT

2

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 ac = 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 / cos ) (sin )

ar = g tan vT2/r

vT = (gr tan )½

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

/g)½ = 2 (L cos /g)½

r

L


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

axis of rotation

Period:t = 2 r / vT =2 (r cot /g)½ = 2 (L cos / g )½ = 2 (5 cos / 9.8 )½ = 4.38 s = 2 (5 cos / 9.8 )½ = 4.36 s = 2 (5 cos 1/ 9.8 )½ = 4.32 s r

L


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 ?

Another example of circular motionLoop-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 = mvB2/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


A horizontal disk is initially at rest and very slowly undergoes constant angular acceleration. A 2 kg puck is located a point 0.5 m away from the axis. At what angular velocity does it slip (assuming aT << ar at that time) if s=0.8 ?

Only one force is in the horizontal direction: static friction

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

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 0.5)1/2

vT = 2 m/s = vT / r = 4 rad/s

Another Example

mpuck= 2 kgS = 0.8 r = 0.5 m g = 10 m/s2

y

xN

mgFs


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?


Lecture 10

Assignment: HW5, (Chapters 7, due 2/24, Wednesday)

For Tuesday: Finish reading Chapter 8 and start Chapter 9

physics 207: lecture 10, pg 1 lecture 10 l goals: exploit newton’s 3 rd law in problems with...

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