TIME OF COMPLETION______________

TIME OF COMPLETION_______________ &nb= sp; NAME_____SOLUTIO= N________________________

<= 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 26, 2006

Total Weight: 100 points

1. = Check your examination for completeness prior to starting. There are a total of nine (9) 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 five (5) multiple choice and four (4) calculation problems. Work all multiple choice problems and 4 calculation problems. Show all work; partial credit will= be given for correct work shown.

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

be located in the classroom.

7. = Start: &= nbsp; 10:30 a.m.

Stop: &= nbsp; 11:45 a.m

&= nbsp; &nbs= p; PROBLEM

&= nbsp; &nbs= p; POINTS

&= nbsp; &nbs= p; CREDIT

&nbs= p; &= nbsp; 1-5

&nbs= p; &= nbsp; 20

&= nbsp;

&nbs= p; &= nbsp; 6

&nbs= p; &= nbsp; 20

&= nbsp;

&nbs= p; &= nbsp; 20

&= nbsp;

&nbs= p; &= nbsp; 8

&nbs= p; &= nbsp; 20

&= nbsp;

&= nbsp; &nbs= p; 9

&= 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. =

1. An ice skater performs a pirouette (a fast spin) by pulling in his outstretched ar= ms close to his body. What happe= ns to his moment of inertia about the axis of rotation?

&nbs= p; a. It does not change.

&nbs= p;

&nbs= p; b. It increases.

(4)

&nbs= p; c. It decreases.

&nbs= p; d. It changes, but it is impossible to tell which way.

2. A wheel with moment of inertia 3.00 kg∙m² has a net torque of 3.50= N∙m applied to it. What angular acceleration does it experience?

&nbs= p; a. 0.857 rad/s²

&nbs= p; b. 1.17 rad/s²

(4)

&nbs= p; c. 3.00 rad/s²

&nbs= p; d. 3.50 rad/s²

3. A fan has blades 0.250 m long rotating at 20.0 rpm. What is the centripetal acceleration of a point on the outer tip of a blade?

&nbs= p; a. 1.10 m/s².

&nbs= p; b. 0.87 m/s².

(4)

&nbs= p; c. 0.55 m/s².

&nbs= p; d. 0.23 m/s².

4. According to Kepler’s first law, Halley’s comet circles the sun in an elliptical path with the Sun at = one focus of the ellipse. What is at the other focus of the ellipse?=

&nbs= p; a. Nothing.

&nbs= p; b. The Earth.

(4)

&nbs= p; c. The comet itself passes through the other focus.

&nbs= p; d. The tail of the comet stays at the other ellipse.

5. The quantity “moment of inertia” = is equivalent, in terms of fundamental quantities of mass, length, and time, t= o:

&nbs= p; a. ML²T^-2.

&nbs= p; b. ML.

(4)

&nbs= p; c. ML².

&nbs= p; d. ML^-1T^-2.

6. A 60.0-kg painter is on a uniform 25-kg scaffold supported from above by ropes. There= is a 4.0-kg pail of paint to one side, as shown. What are the magnitudes of the tension forces exerted by the ropes on the scaffold?

&nbs= p;

(20)

T_Lx =3D T_Lcos (90^{o) =3D
0}

T_Ly =3D T_Lsin (90^{o) =3D
T_L}

T_Rx =3D T_Rcos (90^{o) =3D
0}

T_Ry =3D= T_Rsin (90^o) =3D T_R

w_px =3D w_pcos (-90^{o) =3D
0}

w_py =3D w_psin (-90^{o) =3D
- m_p g =3D - (60.0
kg)(9.81 m/s²) =3D - 589 N}

w_sx =3D w_scos (-90^{o) =3D
0}

w_sy =3D w_ssin (-90^{o) =3D
- m_s g =3D - (25.0=
kg)(9.81
m/s²) =3D - 245 N}

w_{pail x} =3D w_pailcos (-90o) =3D 0

w_{pail y} =3D w_pailsin (-90o) =3D - m_pail g =3D = - (4.00 kg)(9.81 m/s²) =3D - 39.2 N

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

SF_y=3D 0 =3D> T_R + T_LR= 11; 589 N – 245 N – 39.2 N =3D 0

T_{R +
T_L=3D 873 N}= =

Assuming that the axis of rotation is at the point where the left cable is attached to the scaffold:

t_TL =3D 0

t_TR =3D T_R (4.00 m) =

t_wp =3D - (589 N) (2.00 m) =3D - 117= 8 N-m

t_sp =3D - (245 N) (2.00 m) =3D - 490= N-m

t_wp =3D - (39.2 N) (1.00 m) =3D - 39= .2 N-m

St=3D 0

T_R (4.00 m) - 1= 178 N-m – 490 N-m – 39.2 N-m =3D 0

T_{R<=
/sub> =3D
427 N}

T_L=3D&= nbsp; 873 N – 417 N =3D 446 N

<= o:p>

7. A potter’s wheel is rotating around a vertical axis through its center = at an angular velocity of 1.50 rad/s. The wheel can be considered a uniform disk = of mass 5.00 kg and diameter 0.800 m. The potter then throws a 1.00-kg chunk of clay onto the rim of the rotating wheel. What is the angular velocity of the wheel after the clay sticks to it?

I_i =3D ½ m_wR^{2 =3D
½ (5.00 kg)(0.400 m)² =3D 0.400 kg m²}

I_f =3D ½ m_wR^{2 +
m_cR² =3D ½ (5.00 kg)(0.400 m)² + (=
1.00
kg)(0.400 m)² =3D 0.560 kg m²}

I_i w_i =3D I_f w_f

w_f =3D (I_i /I_f) w_i =3D 1.07 rev/s

<= /o:p>

<= o:p>

8. A satellite of mass 5500 kg = orbits the Earth and has a p= eriod of 6200 s.

a. Find the magnitude of the Earth’s gravitational force on the satellite.

Hint: find answer to b first and use it to calculate the gravitational force.

F =3D G m₁ m_2/r²

F =3D G M_s M_{E/R²
=3D (6.673 x 10^-11 Nm²/kg²) (5500 kg) (6.0=
0 x 10²⁴
kg )/(7.26 x 10⁶ m)² =3D 41,77=
9 N}

b.= What is the orbital radius of the satellite?

T² =3D (4p²/GM_E)R³

T²GM_E=3D 4p^R³

R³ =3D T²GM_E/(4p²)

R =3D (T²GM_E/(4p²))^1/3

R =3D ((6200 s)²(6.673 x 10^-11 Nm²/kg²⁼)(6.00 x 10²⁴ kg)/(4p²))^1/3 =3D 7.26 x 10⁶ m

<= o:p>

9. A teenager pushes tangentially on a small hand-driven merry-go-round and is able to accelerate it from rest to an angular velocity of 1.75 rad/s in 10.0 s. Ass= ume the merry-go-round is a uniform disk of radius 2.50 m and has a mass of 760= kg, and two children (each with a mass of 25.0 kg) sit opposite each other on t= he edge.

a. What angular acceleration is req= uired to reach 1.75 rad/s velocity in 10.0 s?

w =3D w₀ + a t

a =3D (w - w₀) / t<= o:p>

a =3D (1.75 rad/s) / (10.0 s) =3D 0.175 rad/s²

c. What is the moment of inertia of= the merry-go-round with two children on it?

I =3D ½ M R² + mR2 + mR² = =3D ½ (760 kg)(2.50 m)² + 2 (25.0 kg)(2.50 m)² =3D= 2688 kg m²

b. Calculate the torque required to produce the acceleration, neglecting frictional torque.

St =3D I a =

St =3D (2688 kgm²)(0.175 rad/s²) =3D 470 N-m

Bonus (5 points): What is the magnitude o= f the tangential force required to produce this torque?

St =3D F R

F =3D St /R =3D (470 N-m)/(2.50 m) =3D 188 N