Solutions for Exam 3 - University Physics I | PHYS 2050, Exams of Physics

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Physics 205H / Exam 3 [Form-A]

Spring 2004

Page 2

“It’s 12:34 am – do you know where your next test question is?”

(50,000 points)

rotating.at the bottom of the ramp if there is no friction and the ball slides withoutan inclined plane 0.400 m above the ground. (a) Find the speed of the ball 1.) A solid ball of mass 0.838 kg and radius 0.135 m starts out on at rest

Easiest method is conservation of energy…

mgh

mv

mgh

mv

mgh

mv

gh

v

v

gh

m s

m

m s

1 2 1

(^12)

2 2 1

(^22)

1 (^2 )

(^22)

1

2 1 (^22)

(^2)

1

2

2 801













;

(.

/

)(.

)

.

/

sliding.(b) Find the speed of the ball at the bottom of the ramp if there is friction and the ball rotates without

I

MR

r v

mgh

mv

I

mgh

mv

I

mgh

mv

I

mv

mr

r v

mgh

mv

mr v r

gh

v

v

v

v

v

v

gh

m s

m

m s

solid ball













HG F

KJI



HG F

KJI





52

2

1 2 1

(^12)

2 1

12

2

2 1

(^22)

2 1

12

1

2 1

(^22)

2 1

(^12)

2 1

(^22)

21 52

2

2

1

2 1

(^22)

21 52

(^2) 2

2

1

2 1 (^22)

5 1 (^22)

(^10 ) (^22) (^10 ) (^22)

(^10 ) (^22)

2

7 10

1

710

2

(. ;

/

)(.

)

.

/

b gc

h

b g

NOTE: 2.368 m/s < 2.801 m/s as expected.

Physics 205H / Exam 3 [Form-A]

Spring 2004

by a taut steel cable at an angle(c) A metal pole 4.00 meters long and a weight 98.1 N is kept from falling

(^)
(^) = 24°

. Draw the Free Body Diagram

forceand the Free Rotation Diagram of the metal pole. There is an unknown

(^) F (^1) from the wall on the base of the metal pole at left – set your axis

of rotation there.

F.B.D.

F.R.D.

(d) Find the tension

(^) T 1 in the steel cable.









T

L

w L

T

w

T

T

w

N

N

y

y

1

(^1)

1

1

0

sin

sin

.

(sin

)

.

b

g

Physics 205H / Exam 3 [Form-A]

Spring 2004

Page 4

(e) The steel cable accidentally snaps. Find the initial angular acceleration

(^)  (^) of the metal pole as it begins

to freely rotate about the left end of the pole. Again, the unknown force

(^) F (^1) from the wall is at the axis of

rotation, so it doesn’t enter into this calculation.

w

mg

m

g w

N

m s

kg

I

mL

kg

m

kg m

w L

I

wL I

N

m

kg m

rad

s

rod end





;

.

(.

/

)

.

.

.

.

.

.

.

.

/

,

3 679^2

31

(^2)

31

(^2)

2

2

2

b

g

b

gb

g

b

gb

g

c

h





OR

w

mg

I

mL

w L

I

mgL

mL

mgL mL

L g

L g

m s

m

rad

s

rod end

;

(.

/

)

.

.

/

,

31

2

31

2

31

(^2)

(^232)

2







c

h

c

h

b

g

Physics 205H / Exam 3 [Form-A]

Spring 2004

One Fish, Two Fish, Red Fish, Star Fish!

IN HONOR OF DR

. (^) SEUSS CENTENNIAL

(^) (50,000 points)

2.) (^)
(a) A plate of mass

(^) m (^) has sides of

(^) 4a (^) and (^) 7a

. Find the center of mass

coordinates

(^) y cm (^) by integrating

(^) y

M

y dm

cm (^)

z

(^1)

, using the

(^) x- (^) and (^) y- axes as

Guessshown.

center

of (^) mass

is (^) in

(^) the

(^) middle,

(^) y cm = +a

^

H G F

K J I

 

H^ GF

K J I





z z

z







a^ M

dm

dy

y

M

y dm

y

M

y dy

M

y dy

M y M a a a a a a a

a

cm cm

a a

a a

a a

aM

(^3)

3

(^2) 3

(^4)

(^2)

(^2)

(^2)

2

2 ; ( ) ( )

c

h

c

h

(c) A torque

(^)

(^) to loosen a bolt consists of a force being applied at a distance of 30.0 cm (0.300 m)

function of angle is given byfrom the axis of rotation. As the bolt gets looser, it gets easier and easier to turn the bolt, so the force as a

F

C

, where C is some constant with

complete revolutions (appropriate units. If the total work done by applying this torque through two

= 2 (^)  (^) radians to 6

(^)  (^) radians) is 1500. J, then find C.





 



HG F

KJ I



z

z

Fd

C

R

W

d

C

Rd

CR

CR

CR

CR

J

C

J

m

J

m

N rad

;

ln

ln(

)

ln(

)

ln

ln

.

.

.

ln

(^26)

2 6

b

g

b

g

Those quasi-units are slippery!

Physics 205H / Exam 3 [Form-A]

Spring 2004

Page 6

(b) A plate of mass

(^) m (^) has sides of

(^) 4a (^) and (^) 7a

. Find the moment of inertia

(^) I

of the plate about the

(^) y- axis as shown, by integrating

(^) I

r dm

z

(^2)

.

This is the corrected version of the problem

.

r

x

a M

dm

dx

I

r dm

x

dx

x dx

x

a

a

a

a

M

a

a

Ma

a a

a a

a a

aM

















 









z

z

z







;

;

(

)

(

)

(^2)

2

(^6)

2

(^6)

(^3) 6

(^3) (^3) (^7) (^3) (^3) (^3)

331

2

c h c h c h

Although not necessary, we could check this result

by (^) using

(^) the

Parallel Axis Theorem and the moment of inertia about

(^) the

(^) center

(^) of

mass.

The new axis is 2.5a = 5L/14 from the c.m.

a

L so

I

M

L

ML

I

I

MD

ML

M

L

ML

ML

ML

ML

ML

PAT

cm

H G F

K J I





HG F

KJ I





331

2

(^14731)

2

(^2)

(^121)

2

2

(^121)

(^2) (^19625)

2

(^58849)

(^58875)

(^2)

588124

(^2)

(^14731)

2

b

g

Physics 205H / Exam 3 [Form-A]

Spring 2004

(d) The wreck of the

(^) RMS Titanic

(^) lies 3821 meters below the surface of the ocean. Find the pressure at

the bottom due to the sea water,

(^)
(^) = 1030 kg/m³

.

P

gh

kg

m

m s

m

Pa

(^3)

2

/

.

/

,

,^ c

hc

hb

g

(e) An object has a rotational motion that follows the following equations. Find the vector acceleration
(^) at time t = 0.

( )

.

(.

/ )

(.

/ )

t

rad

rad

s

rad

s





100

2 00

4 00

(^2)

3

t

t

(^2)

3









( )

.

(.

/

)

(.

/

)

.

(.

/

)

(.

/

)

(.

/

)

(.

/

)

(.

/

)

(.

/

)

(.

/

)

(.

/

)

(.

/

)

(

t

rad

rad

s

rad

s

d td

d td

rad

rad

s

rad

s

rad

s

rad

s

d td

d td

d td

rad

s

rad

s

rad

s

rad

s

rad

s





















0

(^2)

3

(^2)

3

(^2)

3

2 2

(^2)

3

(^2)

3

2

t

t

t

t

t

t

t

t t

(^2)

3

(^2)

3

(^22)

.

/

)

(.

/ ) (. / )

.

/

00

0

3

0

(^2)

3

2

rad

s

rad

s

rad

s

rad s

t

t







b g