TIME OF COMPLETION_______________ &nb= sp; NAME______SOLUTI= ON_______________________

<= span style=3D'font-family:"Times New Roman"'>&nb= sp; = &nb= sp; = DEPARTMENT OF NATURAL SCIENCES

PHYS 1111, Exam 3 = &nb= sp; = &nb= sp; = &nb= sp; = &nb= sp; Section 1

Version 1 &nb= sp; = &nb= sp; = &nb= sp; = &nb= sp; = April 25, 2007

Total Weight: 100 points

1. = Check your examination for completeness prior to starting. There are a total of ten (10) prob= lems on seven (7) pages.

2. = Authorized references include your calculator with calculator handbook, and the Refere= nce Data Pamphlet (provided by your instructor).

3. = You will have 75 minutes to complete the examination.

4. = The total weight of the examination is 100 points.

5. = There are six (6) multiple choice and four (4) calculation problems. Work all cal= culation problems and 5 (five) multiple choice.&nbs= p; Show all work; partial credit will be given for correct work shown.<= o:p>

6. = If you have any questions during the examination, see your instructor who will=

be located in the classroom.

7. = Start: &= nbsp; 6:00 p.m.

Stop: &= nbsp; 7:15 p.m

&= nbsp; &nbs= p; PROBLEM

&= nbsp; &nbs= p; POINTS

&= nbsp; &nbs= p; CREDIT

&nbs= p; &= nbsp; 1-6

&nbs= p; &= nbsp; 20

&= nbsp;

&nbs= p; &= nbsp; 7

&nbs= p; &= nbsp; 20

&= nbsp;

&nbs= p; &= nbsp; 20

&= nbsp;

&nbs= p; &= nbsp; 9

&nbs= p; &= nbsp; 20

&= nbsp;

&= nbsp; &nbs= p; 10

&= nbsp; &nbs= p; 20

&= nbsp; &nbs= p; TOTAL

&= nbsp; &nbs= p; 100

&= nbsp; PERCENTA= GE

CIRCLE THE SINGLE BEST ANSWER FOR ALL MULTIPLE CHOICE QUESTIONS. IN MULTIPLE CHOICE QUESTIONS WHICH REQUIRE A CALCULATION SHOW WORK FOR PARTIAL CREDIT.

Two children are riding on a merry-go-round. Child A is at greater distance from the axis of rotati= on than child B. Which child has the larger angular displacement?

&nb= sp;

a.<= span style=3D'font:7.0pt "Times New Roman"'> Child A.

b.<= span style=3D'font:7.0pt "Times New Roman"'> Child B.

(4)=

c. Both have the same angular displacement.

&nb= sp;

d.<= span style=3D'font:7.0pt "Times New Roman"'> Cannot be determined based on the g= iven information.

&= nbsp;

2. What is the quantity used to measure an object’s resistance to changes in rotational motion?=

&nbs= p; a. Mass.=

&nbs= p; b. Moment of inertia. <= /o:p>

(4)

&nbs= p; c. Angular velocity. =

&nbs= p; d. Angular acceleration.

3. Two forces are applied to a doorknob, perpendicular to the door. The first forc= e is twice as large as the second force. The ratio of the torque of the first fo= rce to the torque of the second is:

&nbs= p; a. 1/4.

&nbs= p; b. 1/2.

(4)

&nbs= p; c. 2.

&nbs= p; d. 4.

4. When you ride a bicycle, what is the direction of the angular velocity of the wh= eels?

&nbs= p; a. To your left.

&nbs= p; b. To your right.

(4)

&nbs= p; c. Forward.

&nbs= p; d. Backward.

5. You sit on a freely rotating stool (no friction). When you extend your arms,=

&nbs= p; a. Your moment of inertia decreases and your angular velocity increases.

&nbs= p; b. Your moment of inertia increases and your angular velocity increases.

(4)

&nbs= p; c. Your moment of inertia decreases and your angular velocity decreases.

&nbs= p; d. Your moment of inertia increases and your angular velocity decreases.

6. Two identical spheres, each of mass M and radius R just touch each other. What = is the magnitude of the gravitational force that they exert on each other?

&nbs= p; a. (GM²)/(R²).

&nbs= p; b. (GM²)/(2R²).

(4)

&nbs= p; c. (GM²)/(4R^=
2).

&nbs= p; d. (2GM²)/(R²).

&n= bsp;

&nb= sp;

= 7. &n= bsp; Shekema and Rodney= are carrying a 40.0 kg block on a= 2.00 m board as shown below. The mass of the board is 5.00 kg. Shekema insists t= hat she is very strong and could have carried the block all by herself, so they place the block 0.500 m away from her and 1.50 m away from Rodney. Find the forces (in Newtons= ) exerted by each to carry the block.

N_Rx =3D N_Rcos (90^{o) =3D
0}

N_Ry =3D N_Rsin (90^{o) =3D
N_R}

N_Sx =3D N_Scos (90^{o) =3D
0}

N_Sy =3D= N_Ssin (90^o) =3D N_S

W_{board x} =3D w_boardcos (-90= ^o) =3D 0

W_{board y} =3D w_boardsin (-90= ^o) =3D - m_board g =3D - (5.00 kg)(9.81 m/s²) =3D - 49.1 N

W_{block x} =3D w_blockcos (-90= ^o) =3D 0

W_{block y} =3D w_blocksin (-90= ^o) =3D - m_block g =3D - (40.0 kg)(9.81 m/s²) =3D - 392 N

SF_x=3D = 0 =3D> 0 =3D 0

SF_y=3D 0 =3D> N_R + N_SR= 11; 589 N – 49.1 N – 392 N =3D 0

N_{R +
N_S=3D 442 N}= =

Assuming that the axis of rotation is at the point where Rodney is holding the board:

t_NR =3D 0

t_NS =3D N_S (2.00 m) =

t_{w board}=3D - (49.1 N) (1.00 m) =3D - 49= .1 N-m

t_{w block} =3D - (392 N) (1.50 m) =3D - 588= N-m

St=3D 0

N_S (2.00 m) = 211; 49.1 N-m – 588 N-m =3D = 0

N_{S<=
/sub> =3D
319 N}

N_R=3D&= nbsp; 442 N – 319 N =3D 123 N

8. A compact disc rotates from res= t to 500 rev/min in 5.50 s.

a. What is its ang= ular acceleration, assuming that it is constant?

= w₀=3D 0

w =3D (500 rev/min)(1 min/60.0 s)(2= p rad/1 rev) =3D 52.4 rad/s

w =3D w₀ + a t

a =3D (w - w0) / t

a =3D (52.4 rad/s) / (5.50 s) =3D 9.52 rad/s^2<=
/sup>

<= /span>

b. &n= bsp; How many revolutio= ns does it make in 5.50 s?

q =3D q₀ + w₀t + 1/2 a t² =3D ½ (52= .4 rad/s²)(5.50 s)² =3D 793 ra= d =3D 126 rev

<= /span>

c. &n= bsp; At the moment the disk reaches the angular velocity of 500 rev/min, what is the tangential velocity of a point on the rim of the disc = 6.00 cm from the center of rotation ?

V_t =3D = w R =3D (52.3 rad/s= )(0.0600 m) =3D 3.14 m/s

9.= Penny goes bowling with her friends. She notices that a bowling ball of radius 11= .0 cm and mass m =3D 7.20 kg is rolling without slipping on a horizontal ball = return at 2.00 m/s. It then rolls without slipping up a hill to a height h before momentarily coming to rest. Find h.

E_i =3D E_f

m g y_i + ½ m V_i² + ½ I w_i² _{=3D m
g y_f + ½ m V_f²+ ½ I w_f²}

y_{i<=
/sub> =3D
0}

V_{i<=
/sub> =3D 2.00
m/s}

&nb= sp;

w_i =3D V_i /R

&nb= sp;

y_{f<=
/sub> =3D h}

&nb= sp;

V_{f<=
/sub> =3D 0}

&nb= sp;

w_f =3D 0

½ m V_i² + ½ I w_i² _{=3D m
g y_f}

m V_i² + = I w_i² _{=3D 2 m
g y_f}

m V_i² + = (2/5 m R²)(V_i /R)² =3D 2 m g y_f ₌

V_i² + (2/5) V<= sub>i² =3D 2 g y_f

(7/10) V_i² =3D g y_f

y_f =3D 0.285 m

10. Amanda, David, and Rota= sha play with two children who want a ride on a merry-go-round. The merry-go-round is a solid disc= that rotates freely(without friction) around its center and has a moment of inertia of 500 kg-m²relative to this axis. The children have the same masses of 25.0 kg = and sit at distances of 1.00 m and 2.00 m away from the center of merry-go-roun= d. The radius of marry-go-round is 3.00 m.

a.<= span style=3D'font:7.0pt "Times New Roman"'> What is the moment of inertia of the merry-go-round with the children on it relative to the center of rotation?<= o:p>

I =3D 50= 0 kg-m²+(25.0 kg)(1.00 m)² + (25.0 kg)(2.00 m)² =3D 625 = kg-m²

b.<= span style=3D'font:7.0pt "Times New Roman"'> Amanda, David, and Rotasha pull the merry-go-round with tangential forces as shown in the figure. (The magnitud= es are: A =3D 300 N, D =3D 150 N, R =3D 200 N)

What is the net torque = that they exert on the merry-go-round?

St =3D (300 N)(3.00 m) + (200 N)(3.00 m) – (150 N)(3.00 m) =3D 1050 N-m

c.<= span style=3D'font:7.0pt "Times New Roman"'> What is the angular acceleration of= the merry-go-round?

a =3D St/I =3D (1050 N-m)/(625 kg-m^{2) =3D 1.68 rad/s²}

&nb= sp;

q =3D q0 + w0 t + 1/2 a t2 =3D ½ (52= .4 rad/s2)(5.50 s)2 =3D 793 ra= d =3D 126 rev

9.= Penny goes bowling with her friends. She notices that a bowling ball of radius 11= .0 cm and mass m =3D 7.20 kg is rolling without slipping on a horizontal ball = return at 2.00 m/s. It then rolls without slipping up a hill to a height h before momentarily coming to rest. Find h.

q =3D q₀ + w₀t + 1/2 a t² =3D ½ (52= .4 rad/s²)(5.50 s)² =3D 793 ra= d =3D 126 rev