Engineering Mechanics - PKRB: Work and Energy - Discussion
Discussion Forum : PKRB: Work and Energy - General Questions (Q.No. 1)
1.
A man having a weight of 180 lb sits in a chair of the Ferris wheel, which has a weight of 15,000 lb and a radius of gyration of ko = 37 ft. If a torque of M = 80(103) lb • ft is applied about O, determine the angular velocity of the wheel after it has rotated 180°. Neglect the weight of the chairs and note that the man remains in an upright position as the wheel rotates. The wheel starts from rest in the position shown.
= 0.888 rad/s = 0.836 rad/s = 0.874 rad/s = 0.849 rad/sDiscussion:
1 comments Page 1 of 1.
Jorge Oberdan said:
1 decade ago
M = 8,0* 10^4 lb.ft
1,5 *10^4 lb e 6,0.10ft
T1 + ∑U1,2 = T2
8,0 . 10 ^4(π) – (180)*(120)=
=(1/2){[(1,5 * 10^4/3,22 * 10)(3,7*10)²]ω² + (1/2)[(1,8 * 10²/3,22*10)](60ω)²}=
ω = 0,836 rad/s
1,5 *10^4 lb e 6,0.10ft
T1 + ∑U1,2 = T2
8,0 . 10 ^4(π) – (180)*(120)=
=(1/2){[(1,5 * 10^4/3,22 * 10)(3,7*10)²]ω² + (1/2)[(1,8 * 10²/3,22*10)](60ω)²}=
ω = 0,836 rad/s
(1)
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