1. d_theta = 2*PI, d_t = 60 sec --> omega = d_theta / d_t = 0.105 rad/s 2. omega = alpha * t --> t = 120/4.0 = 30 s 3. theta = omega * t + 1/2 * alpha * t^2 = 1800 rad since alpha = 14.0 rad/s^2 4. omega = 240rev/min = 240*2*PI/60 = 25.13 rad/s --> v = r*omega = 12.5 m/s 5. torque = r*F*sin(theta) = 0.2*(70*9.81) = 137 N.m 6. torque = I * alpha --> I = torque/alpha = 960/6.20 = 2/3 * MR^2 --> M = 960/6.20 / (2/3) / 1.9^2 = 64.4 kg 7. L = 2 * r*mv = 2 * .5 * 3 * 2 = 6.0 8. omega = 2*PI/(24*60*60) = 7.2722x10^(-5) --> L = m*r^2*omega = 84*(6370*10^3)^2*7.2722x10^(-5) = 2.5x10^11 9. L_i = L_f --> I_i * omega_i = I_f * omega_f --> omega_f = 6.0*1.2/2.0 = 3.6 rev/s 10. K.E. = 1/2 * I * omega^2 --> KE_i < KE_f
Solutions (Exam 2.)
1. a = v^2 / r = v^2 / (R_E + h) = 8.38 m/s^2
2. centfugal force of m = weight of M
m*v^2 / r = Mg --> v = (Mgr/m)^(-1/2) = 5.42 m/s
3. v = a*t, a_av = (50-0)/2 / 10.0 = 2.5 m/s^2 --> v = a_av * t = 10 m/s
4. f = mu * N, N = m*g + F*sin(40) = 43.94 [N] --> f = 0.25*43.94 = 11 [N]
5. a = F_x / m = (F*cos(40) - f) / 3.5 = 0.14 m/s^2
6. (change of potential energy of block) = (potential energy of compressed spring)
m*g*del_h = 1/2 * k * x^2 --> 2.0*9.81*0.4 = 1/2 * 1960 * x^2
--> x = 10 cm
7. x_CM = (m_1 * x_1 + m_2 * x_2 + m_3 * x_3) / (m_1 + m_2 + m_3) = 1.1
y_CM = (m_1 * y_1 + m_2 * y_2 + m_3 * y_3) / (m_1 + m_2 + m_3) = 1.3
8. By the conservation of linear momentum, (m: mass of bullet, M1: 1st block, M2: 2nd)
M1 * v1 + (M2+m) * v2 = m * v --> v = 937 m/s
9. Right after the bullet emerges from the 1st block, the bullet and the 1st block are moving.
Again, by the momentum conservation,
m * v = M1 * v1 + m * v' --> v' = 721 m/s
10-11. By the momentum conservation along x and y axis (m: mass of the ball)
x: m * 2.2 = m * 1.1 * cos(60) + m * v * cos(theta) ....eq. 1
y: m * 1.1 * sin(60) = m * v * sin(theta) .........eq. 2
From eq. 2,
v = 1.1 * sin(60) / sin(theta) .... eq. 3
Substituting eq. 3 into eq. 1,
sin(theta)/cos(theta) = tan(theta) = (1.1 * sin(60)) / (2.2 - 1.1 * cos(60))
---> theta = 30 degree
---> v = 1.9 m/s
12. elastic since K.E. was conserved.
13-15. Refer to Ex. 7-5 of our textbook.
* Final : 8-11am on May 6 : Ch. 1 -- 11
* 3rd Test : April 24 (THURSDAY) : Ch. 10 & Ch. 11 (sections 1, 6, 7)
* Study Prob.
Ch. 11 .... Probs. #2, #13, #49, #61
* Homework Set #10
Ch. 10 .... Probs. #9, #14, #20, #25, #28, #42 (Due Apr. 24, 2003)
* Homework Set #9
Ch. 9 .... Probs. #3, #13, #19, #25, #31 (Due Apr. 17, 2003)
* 2nd Test : April 10 (THURSDAY) : Ch. 6 - 9
* Homework Set #8
Ch. 8 .... Probs. #3, #10, #20, #25, #66 (Due Apr. 3, 2003)
* Homework Set #7
Ch. 7 .... Probs. #23, #37, #44, #54, #67 (Due Apr. 2, 2003)
* Homework Set #6
Ch. 6 .... Probs. #7, #19, #26, #30, #42 (Due Mar. 25, 2003)
* Homework Set #5
Ch. 5 .... Probs. #9, #15, #16, #22, #32 (Due Mar. 11, 2003)
* 1st Test : Feb. 27
* Homework Set #4
Ch. 4 .... Probs. #6, #14, #25, #38, #57 (Due Feb. 20, 2003)
* Homework Set #3
Ch. 3 .... Probs. #6, #20, #32, #39, #49 (Due Feb. 11, 2003)
* Homework Set #2
Ch. 2 .... Probs. #17, #23, #33, #39, #66 (Due Feb. 4, 2003)
* Homework Set #1
Ch. 1 .... Prob. #2, #5, #14, #23, #36 (Due Jan. 30, 2003)
* Homework Grader : Fei Yen (fyen@uh.edu, S&R1 426h)
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Physics 1301
Syllabus -- Spring 2003
------------------------------------------------------------------------------
Instructor: Choong-seop Lee
Office: Rm 619A
Office Hours: 4pm-5pm (TTH) (Otherwise, appointment-based only please!)
Phone: 3-3594
E-mail: clee1@uh.edu
Homework grader: Yen Fei
Text Author: James S. Walker (Covering Ch. 1 - 18)
Determination of Grade
Tests (3) 600 pts
Homework 100 pts
Final 500 pts
--------------------------------------------
Total 1200 pts
A Typical Grade Distribution
A 100 -- 87%
B 86 -- 78%
C 77 -- 69%
D 68 -- 60%
1) Homework will be assigned on each chapter. You are encouraged to discuss the
problems with other students in the class. The best 15 (out of 18) homeworks will
be counted.
2) There are NO make-up tests.