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Unformatted text preview: 1 Solutions to Practice test 4 1. A bicycle is moving forward at 4.5 m.s1 . 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 ms1 , r = 0.75/2 = 0.375 m. 11 4.5 ms 12 rad.s 0.375 m v r ϖ = = = a. 12 rad.s1 2. What is the angular velocity of a disc rotating at 120 rpm in rad/s? ( 291 120 2 120 rpm 4 rad.s 60 rad s π π × = = d. 4 π 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? ϖ o = 0, ϖ f = 120 rpm = 4 π rad.s1 , t = 20 s 12 4 rad.s rad.s 20 s 5 f o t ϖ ϖ π π α = = = c. π /5 rad.s2 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 τ = 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 22 5 m.N 25 rad.s , 5 . ; I = kg.m 5 ( /5) rad.s m N π τ α τ α π π = = = = a. 25/ π kg ⋅ m 2 6. What is the moment of inertia of a 10kg 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 τ = 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 , τ = 30 m.N 2 2 30 m.N = 273 rad.s 0.11 kg.m I τ α = = b. 273 rad.s2 9. What is the angular velocity of this disc in question 8 after 1.2 s? ϖ o = 0, α = 273 rad.s2, t = 1.2 s; ϖ = ϖ o + α t = 0 + 273 rad.s2 × 1.2 s = 328 rad.s1 . a. 328 rad.s1 10. What is the approximate linear velocity of a point on the rim of the wheel in question 9 after 1.2 s? ϖ = 328 rad.s1 , r = 0.15 m; v = r ϖ = 0.15 m × 328 rad.s1 = 49 ms1 . d. 49 m.s1 11. What is the angular momentum of the disc after 1.2 s? I = 0.11 kg.m 2 , ϖ = 328 rad.s1 ; L = I ϖ = 0.11kgm 2 × 328 rad s1 = 36.1 kgm 2 s1 a. 36.1 kgm 2 s1 12. What is the angular displacement of the disc in 1.2 s θ = ϖ o t + ½ α t 2 = 0 + ½ × 273 rad.s2 × 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 π radians. 196.6 rad Number of revolutions = 31.3 2 2 rad θ π π = = 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 , ϖ = 328 rad.s1 ; 2 2 1 1 0.11 328 5917 J 2 2 K I ϖ = = × × = c. 5917 J 15. A wheel rotating at 55 rad.s1 comes to a stop in 11 seconds. What is the angular acceleration of the wheel? ϖ o = 55 rad.s1 , ϖ f = 0, t = 11 s; 2 55 5 rad.s 11 f o t ϖ ϖ α = = =  a. –5 rad.s2 16. What is the angular displacement of the wheel while coming to rest?...
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 Fall '10
 George
 Physics, Energy, Force, Mass, kg

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