THE ANGULAR SPEED OF A MOTOR WHEEL IS INCREASED FROM 1200rpm to 3120 rpm in 16 seconds (1) what is its angular acceleration,assuming the acceleration to be uniform (2) how many revolutions does the wheel make during this time?
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Answered by
234
use w1=w2+at
w1=initial angular speed
w2=final angular speed
w1=1200rpm=20rps
w2=3120rpm=52rps
52=20+a*16
a=2
then for no. of revolutions use s=w1t+(1/2)a*t^2
s=20*16+(1/2)*2*16^2
no. of revolutions = 576
w1=initial angular speed
w2=final angular speed
w1=1200rpm=20rps
w2=3120rpm=52rps
52=20+a*16
a=2
then for no. of revolutions use s=w1t+(1/2)a*t^2
s=20*16+(1/2)*2*16^2
no. of revolutions = 576
Answered by
196
ω₁ = 1,200 rpm = 1,200/60 = 20 rps = 20 * 2π = 40π rad/sec
ω₂ = 3,120 rpm = 3,120 *2π/60 = 104π rad/sec
duration = t = 16 sec
Angular acceleration α is assumed constant.
α = Δω/ t = (ω₂ - ω₁) / t = 64π/16 = 4π rad/sec²
θ = angle rotated = ω₁ t + 1/2 α t²
= 40π * 16 + 1/2 * 4π * 16² rad
= 1152 π rad
Number of rotations turned in the 16 sec = θ/2π
= 1152π/2π = 576
ω₂ = 3,120 rpm = 3,120 *2π/60 = 104π rad/sec
duration = t = 16 sec
Angular acceleration α is assumed constant.
α = Δω/ t = (ω₂ - ω₁) / t = 64π/16 = 4π rad/sec²
θ = angle rotated = ω₁ t + 1/2 α t²
= 40π * 16 + 1/2 * 4π * 16² rad
= 1152 π rad
Number of rotations turned in the 16 sec = θ/2π
= 1152π/2π = 576
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