Practice_Test_4_Answers

Practice_Test_4_Answers - 1 Solutions to Practice test 4 1....

Info iconThis preview shows pages 1–4. 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

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: 1 Solutions to Practice test 4 1. A bicycle is moving forward at 4.5 m.s-1 . If the diameter of the wheel is 0.75 m, what is the angular velocity of a point on the rim of the wheel? v = 4 ms-1 , r = 0.75/2 = 0.375 m. -1-1 4.5 ms 12 rad.s 0.375 m v r = = = a. 12 rad.s-1 2. What is the angular velocity of a disc rotating at 120 rpm in rad/s? ( 29-1 120 2 120 rpm 4 rad.s 60 rad s = = d. 4 3. A tangential force of 10 N is applied to a disc initially at rest and it attains an angular velocity of 120 rpm in 20 s. What is the angular acceleration of the disc? o = 0, f = 120 rpm = 4 rad.s-1 , t = 20 s -1-2 4 rad.s rad.s 20 s 5 f o t - = = = c. /5 rad.s-2 4. If the radius of the wheel in question 3 is 0.5 m, what is the torque on the wheel? Tangential force = 10 N, r = 0.5 m = r F = 0.5 m 10 N = 5 m.N b. 5 m.N 5. What is the moment of inertia of the disc in question 3? -2 2-2 5 m.N 25 rad.s , 5 . ; I = kg.m 5 ( /5) rad.s m N = = = = a. 25/ kg m 2 6. What is the moment of inertia of a 10-kg disc of radius 15 cm? 2 2 2 1 1 I = Mr = 10 kg (0.15 m) 0.11 kg.m 2 2 = c 0.11 kg m 2 2 7. If a tangential force of 200 N is applied to the disc in question 6, what is the torque it produces? Tangential force = 200 N, r = 0.15 m = r F = 0.15 m 200 N = 30 m.N c. 30 m.N 8. What is the angular acceleration of the disc in question 7? I = 0.11 kgm 2 , = 30 m.N -2 2 30 m.N = 273 rad.s 0.11 kg.m I = = b. 273 rad.s-2 9. What is the angular velocity of this disc in question 8 after 1.2 s? o = 0, = 273 rad.s-2, t = 1.2 s; = o + t = 0 + 273 rad.s-2 1.2 s = 328 rad.s-1 . a. 328 rad.s-1 10. What is the approximate linear velocity of a point on the rim of the wheel in question 9 after 1.2 s? = 328 rad.s-1 , r = 0.15 m; v = r = 0.15 m 328 rad.s-1 = 49 ms-1 . d. 49 m.s-1 11. What is the angular momentum of the disc after 1.2 s? I = 0.11 kg.m 2 , = 328 rad.s-1 ; L = I = 0.11kgm 2 328 rad s-1 = 36.1 kgm 2 s-1 a. 36.1 kgm 2 s-1 12. What is the angular displacement of the disc in 1.2 s = o t + t 2 = 0 + 273 rad.s-2 1.2 2 = 196.6 rad a. 196.6 rad 13. What is the number of revolutions of the disc in 1.2 s? One revolution corresponds to an angular displacement of 2 radians. 196.6 rad Number of revolutions = 31.3 2 2 rad = = b. 31.3 3 14. What is the kinetic energy of rotation of the disc after 1.2 s? I = 0.11 kg.m 2 , = 328 rad.s-1 ; 2 2 1 1 0.11 328 5917 J 2 2 K I = = = c. 5917 J 15. A wheel rotating at 55 rad.s-1 comes to a stop in 11 seconds. What is the angular acceleration of the wheel? o = 55 rad.s-1 , f = 0, t = 11 s; -2 55 5 rad.s 11 f o t -- = = = - a. 5 rad.s-2 16. What is the angular displacement of the wheel while coming to rest?...
View Full Document

Page1 / 25

Practice_Test_4_Answers - 1 Solutions to Practice test 4 1....

This preview shows document pages 1 - 4. Sign up to view the full document.

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