Circular MotionHard
Question
A particle moves along a circle of radius 
m with tangential acceleration of constant magnitude. If the speed of the particle is 80 m/s at the end of the second revolution after motion has begun, the tangential acceleration is:
m with tangential acceleration of constant magnitude. If the speed of the particle is 80 m/s at the end of the second revolution after motion has begun, the tangential acceleration is:Options
A.160 π m/s2
B.40 π m/s2
C.40 m/s2
D.640 π m/s2
Solution
r = 
m, at = constant
n = 2nd revolution
v = 80 m/s
ω0 = 0, ωf =
= 
= 4μ rad/sec
θ = 2π × 2 = 4π
from 3rd equation
ω2 = ω02 + 2αθ
⇒ (4π)2 = 02 + 2 × α × (4π)
α = 2π rad/s2
at = αr = 2π ×
= 40 m/s2 Ans.
m, at = constant n = 2nd revolution
v = 80 m/s
ω0 = 0, ωf =
= = 4μ rad/secθ = 2π × 2 = 4π
from 3rd equation
ω2 = ω02 + 2αθ
⇒ (4π)2 = 02 + 2 × α × (4π)
α = 2π rad/s2
at = αr = 2π ×
= 40 m/s2 Ans.Create a free account to view solution
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