Week5exercises - from text PDF

Preview

Summary

Final exam, with the work shown. Given at the end of the course...

Description

Charles Terrill Southern New Hampshire University Module 5 Problem Set – 3.4 (2,16); Section 3.5 (2) 3.4 2. The force due to gravity is: F1=400 lb The force due to air resistance is: F2 =−10 v

The net force acting on the object is: F =F1 + F 2=400−10 v The force does not depend on x, enabling us to regard the above equation as an FOE.

m

dv =F (t , v ) dt

Then,

F1=mg→ m=

F 1 400 = =12.5 slug g 32

Now we apply newton’s second law:

m

dv =400−100 v dt

Integrate:

12.5

dv =400−10 v dt

dv =32−0.8 v dt

1 dv=−dt 0.8 v−32

1

∫ .8 v−32 dv=−∫ dt 5 ln |.8 v− 32|=−t +C1 4 ln|.8 v −32|=−.8 t +C2 −0.8t

0.8 v −32=e

+ C2

−.8 t

0.8 v −32=C 3 e

t 0.8 v =32+ C−.8 3e

−.8t

v (t ) =40+ C e

Using substitution value condition, we find the value of C: 0

0=40+C e

−40 =C

0=40+C

Therefore: v (t ) =40 −40 e−.8 t Integrate: x (t ) =∫ v ( t ) dt=∫ ( 40−20 e

−.8t

) dt

x (t ) =40 t+50 e−.8 t + C Substitute the initial value condition x(0) = 0 and obtain the value of C: 0=40 ( 0) + 50 e 0 + C

0=50+C

−50=C

−.8 t

x (t )=40 t+50 e

−50

According to the problem the object hits the ground at 500:

500=40 t−50 ( 1−e−.8t )

500=40 t−50

500 + 50=40 t

−.8t

500=40 t+50 e

t=

−50

500+ 50 40

t ≈ 13.75 sec

16. Notation: T – Motor Torque ω – angular velocity I – inertia ω1 =initial angular velocity

When friction is the reason torque is slowing or ceasing and is proportional to the angular velocity we use: T 1 =K √ω K is proportionally constant. Use the Torque formula: Moment of inertia x angular acceleration = sum of the torques we get:

I

dw =T −K √ ω dt

We use separation of variables in order to solve the differential equation:

Id ω =dt T −K √ ω

√ ω−

⇒−

2 I √ ω − 2TIln|T −K √ ω | =t+C1 K K2

TIln |T −K √ ω| =t +C1 K 2I ⇒ ¿ K

⇒K √ ω−Tln|T −K √ ω|=

⇒√ ω−

2

K t +C 2I

TIln|T −K √ ω| Kt = +C 2 K 2I

C=K √ ω0−Tln∨T − K √ω0 ∨¿

K √ω−Tln|T −K √ ω|=K √ ω 0−Tln|T −K √ ω0|−

K2t 2I

Section 3.5 (2) 2. I used equation 3 on page 119 which is a general equation. e 100t

sin 100 t dt + K 0.000001 ∫¿ −100 t ¿ I (t )=e

100 t 100 sin 100 t−100 cos ¿ ¿ t e 100 ¿ 100 ¿ −100t ¿e ¿

¿

sin 100 t−cos 100 t +K e−100 t 2

For I(0)= 0, we obtain -1.2+K=0 so K=1/2 and the current is : sin 100 t−cos 100 t+e−100 t I ( 0 )=.5 ¿

The resistor and inductor voltages are given by E R ( t) =RI ( t ) =I ( t )

E L ( t) =L

dI =( 0.5)( cos 100t +sin 100 t−e−100t ) dt

Applying Kirchoff’s Voltage law yields: RI +q∨C=e (t) The capacitor is the rate of change of its charge: I= dq/dt I followed the steps in the second example but when I look up the answer to number one in the book, I see mine isn’t looking similar. Not sure what I am doing in the first place just trying to figure it out step by step....