mech 2241 (w2015) lecture 08

Stephen G. [email protected]

Office: Room 202 SW9(604) 432‐8332

Lecture 8

MECH 2241 (W2015) Dynamics S. Ribarits / D. Stropky Week 4/Lecture 8/Slide 2 of 18

∆∆

∆∆∆∆

∆∆∆∆

Clockwise(CW)

CounterClockwise

(CW)

∆∆

∆∆

∆∆

∆∆∆∆

1212

22

1212

22

Averages

Constant Acceleration Equations

Averages

Constant Acceleration Equations

Relations


NORMAL AND TANGENTIAL ACCELERATION

Section 16‐3



Why don't people on the Roundup ride fall out when they are at the top?

There must be some invisible force that is counteracting gravity!

We will learn that a free particle (i.e., a human) has to be accelerated to feel a force.

But we already learned that angular acceleration is given by:


,


Let's look at rotationalmotion in detail

A particle is traveling in a circle of radius . Imagine the motion as, at every instant, a rope attached to the particle is pulling it tangential to the circle. If the rope has constant speed

, then we already know that the angular velocity of the particle is,

We also know that if the imaginary rope is pulled with constant acceleration (meaning the speed would be increasing or decreasing), that the angular acceleration of the particle is,


constant


,

Since the acceleration depicted in the previous slide is tangential to the circle, let us denote it with a subscript ,

Now let's reconsider the case where the imaginary rope is pulled at constant speed ,

0

0 constant

tangentialacceleration



We know from the Roundup ride discussion that there must be an acceleration in the constant speedcase. So where does it come from?

= = constant= =

v

Let us look at the particle at another position on the circle.

v

The speed at points and is equal, however the direction is not.

It is the change in direction that is causing the acceleration. From our previous work we know that,

∆∆

∆∆



, constant= =

Let us look in detail at the change in velocity, ∆ , over a very small change in angle ∆ .

∆∆

∆∆

∆

The direction of is toward the center of the circle.



∆∆∆∆

∆

∆

We know that because the velocity magnitude is not changing. Let ,

∆ 2 cos ∆

but we know,

For a very small angle ∆ ,

to circle center

∆ 2 1 cos ∆

∆ 2 1 cos ∆

∆ 2 1 cos ∆

cos ∆ ≅ 1∆2

∆ in radians!



∆∆∆∆

∆

∆

∆ 2 1 cos ∆

∆∆

cos ∆ ≅ 1∆2

∆ 2 1 1∆2

∆ ∆

∆ ∆

∆ ∆

∆∆∆∆

but

normalacceleration



,

= tangential acceleration

This is the acceleration due to a change in speed around the circle.

r tangentialacceleration

= normal acceleration

This is the acceleration due to a change in direction around the circle.

normalacceleration



1. As before, pay careful attention to sign convention of both given data and calculated values. If initial velocity or displacement is positive, then acceleration is positive and deceleration is negative.

2. Mating pulleys and gears have a common tangential velocity.

3. For a rotating object:a) If angular velocity is zero, but may

be present.b) If angular acceleration is zero, but

may be present.



Example11‐10

A 1.2 m wheel, mounted in a vertical plane, accelerates uniformly from rest at 3 rad/s2 for 5 seconds and then maintains uniform velocity in a clockwise direction. Determine the normal and tangential acceleration of a point at the top of the wheel for t=0 seconds and t=6 seconds.

At seconds: 0 0

3 3 0.6 1.8

At seconds: 0 0

0 3

speed increase only to 5 seconds!

15

15 0.6 135



Problem 11‐65Point A has a total acceleration of 240 in/s2

For member ABC, rotating clockwise, determine:a) angular velocity b) angular acceleration

35

35240

Total acceleration:

sin 35°

cos 35°

tan512 22.6°

5 12 13



35

240 sin 35° 22.6°

240 cos 35° 22.6°

51.5 234.4

234.4

13

51.5

13

4.254.25

3.963.96


Relative Motion on a plane


ANGULAR MOTION KINEMATICS

1. Determine the normal and tangential accelerations for point P in the figure.(Answers an = 270 m/sec2 →, at = 1.5 m/sec2 ↓)

2. The object shown in the figure rotates about A and accelerates from an initial speed of 10 rpm to 40 rpm in 5 seconds. At t = 5 seconds, determine the total acceleration of point B for the position shown.(Answer aB = 87.8 m/sec2 ∠ ‒55.2°)


ANGULAR MOTION KINEMATICS

3. A car accelerates uniformly from rest to 50 mph over a distance of 1/8 mile along a curve of 700 ft radius. Determine the total acceleration at the instant that the speed of 50 mph is reached.(Answer 8.69 ft/sec2)

4. The total acceleration of a point A is 4 m/sec2∠ 45°. For member ABCD determine (a) the angular velocity and (b) the angular acceleration . (Answer = 3.3 rad/sec CCW, = 2.17 rad/sec2 CW)

mech 2241 (w2015) lecture 08

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