A bug on the rim of a 10.0 in. diameter disk if the disk moves from rest to an angular speed of 75rev/min in 4.0 s? one second after the bug starts from rest, what is the total acceleration?
Answers
Answered by
10
ngular speed , ω = 75 rev/min
we should first change it in rad/sec
e.g., ω = 75 × 2π/60 rad/sec = 7.85 rad/sec
Now, angular acceleration , α = ω/t [ ∵ bug moves from rest so, ω₀ = 0]
so, α = 7.85/4 = 1.9625 rad/sec² [ ∵ t = 4sec ]
now, we have to find out tangential acceleration at t = 1 [ ∵one second after bug starts from rest ]
at = αr , here r is radius of circular path
r = 10/2 inch = 5 × 0.0254 m = 0.127 m
So, tangential acceleration, a = 0.127 × 1.9625 m/s²
= 0.249 ≈ 0.25 m/s²
Now, find out centripetal acceleration , ac
ac = ω²r
ω = αt = 1.9625 rad/s² × 1s = 1.9625 rad/s
so, ac = (1.9625)² × 0.127 m = 0.489 m/s²
Now, total acceleration = √(at² + ac²)
= √{(0.25)² + (0.489)²}
=0.5486 m/s²
we should first change it in rad/sec
e.g., ω = 75 × 2π/60 rad/sec = 7.85 rad/sec
Now, angular acceleration , α = ω/t [ ∵ bug moves from rest so, ω₀ = 0]
so, α = 7.85/4 = 1.9625 rad/sec² [ ∵ t = 4sec ]
now, we have to find out tangential acceleration at t = 1 [ ∵one second after bug starts from rest ]
at = αr , here r is radius of circular path
r = 10/2 inch = 5 × 0.0254 m = 0.127 m
So, tangential acceleration, a = 0.127 × 1.9625 m/s²
= 0.249 ≈ 0.25 m/s²
Now, find out centripetal acceleration , ac
ac = ω²r
ω = αt = 1.9625 rad/s² × 1s = 1.9625 rad/s
so, ac = (1.9625)² × 0.127 m = 0.489 m/s²
Now, total acceleration = √(at² + ac²)
= √{(0.25)² + (0.489)²}
=0.5486 m/s²
Similar questions