11homework09 - Husain, Zeena Homework 9 Due: Oct 28 2003,...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Husain, Zeena Homework 9 Due: Oct 28 2003, 4:00 am Inst: H L Berk 1 This print-out should have 22 questions, check that it is complete. Multiple-choice questions may continue on the next column or page: find all choices before making your selection. The due time is Central time. 001 (part 1 of 3) 10 points Given: g = 9 . 8 m / s 2 . Two pulley wheels, one of radius 0 . 015 m, the other of radius 0 . 03 m, are mounted rigidly on a common axle, as in the figure. The rotational inertia of the two pulleys, which are clamped together, is 1 . 7 kg m 2 . A(n) 1 . 3 kg mass is attached on the left and a(n) 1 . 8 kg mass on the right, as shown. 1 2 m R 1 m R 2 Find the angular acceleration of the system. Take clockwise direction as positive. Correct answer: 0 . 198659 rad / s 2 . Explanation: 1 2 m g R 1 m g R 2 T 1 2 T 2 T T 1 Lets assume m 2 moves downward and take motion downward as positive for m 2 and mo- tion upward as positive for m 1 , take clockwise as positive for the pulley wheels. Since the two masses are connected to the pulleys, their accelerations are a 1 = R 1 and a 2 = R 2 . Applying Newtons second law to m 1 , m 2 , and the pulleys separately, we obtain T 1- m 1 g = m 1 a 1 = m 1 R 1 (1) m 2 g- T 2 = m 2 a 2 = m 2 R 2 (2) net = T 2 R 2- T 1 R 1 = I . (3) Solving these equations, we find = m 2 g R 2- m 1 g R 1 I + m 1 R 2 1 + m 2 R 2 2 = 0 . 198659 rad / s 2 . The fact that > 0 indicates that our pre- vious assumption that m 2 moves downward is right. If < 0, our assumption was not correct and m 2 would move upward, but the equations obtained to calculate T 1 and T 2 would be still correct and we need not reas- sume the direction of the motion and do it again. 002 (part 2 of 3) 10 points Find the tension in the cord attached to the 1 . 3 kg mass. Correct answer: 12 . 7439 N. Explanation: T 1 = m 1 ( g + R 1 ) = 12 . 7439 N . 003 (part 3 of 3) 10 points Find the tension in the cord attached to the 1 . 8 kg mass. Correct answer: 17 . 6293 N. Explanation: T 2 = m 2 ( g- R 2 ) = 17 . 6293 N . 004 (part 1 of 3) 10 points Given: g = 9 . 8 m / s 2 . A thin cylindrical shell and a solid cylinder have the same mass and radius. The two are released side by side and roll down, without slipping, from the top of an inclined plane that is 3 . 4 m above the ground. Husain, Zeena Homework 9 Due: Oct 28 2003, 4:00 am Inst: H L Berk 2 Find the final linear velocity of the thin cylindrical shell. Correct answer: 5 . 77235 m / s. Explanation: Basic Concept: The mechanical energy is conserved K rot + K tran = U i . H S h 1 2 S Solution: Because there is no slipping, v 1 = 1 R The rotational inertia of the cylindrical shell is I 1 = m R 2 Thus, from conservation of energy 1 2 I 1 2 1 + 1 2 m ( 1 R ) 2 = m g H (1) 1 2 ( m R 2 ) v 1 R 2 + 1 2 m v 2 1 = m g H ....
View Full Document

Page1 / 8

11homework09 - Husain, Zeena Homework 9 Due: Oct 28 2003,...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online