A boy is playing with a tire of radius 0.5m. He accelerates it from 5r...
r=0.5m, t=15s
n1=5rpm=5/60 rps
n2=25rpm=25/60rps
… ω1=2π(5) rad/s
… ω2=2π(25) rad/s
As,
a=r∝ [∝=ω2- ω1/t]
so, a=0.5[2π(n2-n1)/t]
a=0.5x6.28x20/60x15
a=6.28x2/60x3
a=6.28/90
a=0.697 m/s2
a≈0.7 m/s2
A boy is playing with a tire of radius 0.5m. He accelerates it from 5r...
Given, radius of the tire, r = 0.5 m
Initial angular velocity, ω1 = 5 rpm
Final angular velocity, ω2 = 25 rpm
Time taken, t = 15 seconds
We need to find the linear acceleration of the tire.
Formula used:
Linear velocity, v = rω
Angular acceleration, α = (ω2 - ω1)/t
Linear acceleration, a = rα
Calculation:
Convert rpm to rad/s: ω = 2πn/60, where n is rpm
Initial angular velocity, ω1 = (2π × 5)/60 = 0.524 rad/s
Final angular velocity, ω2 = (2π × 25)/60 = 2.618 rad/s
Angular acceleration, α = (ω2 - ω1)/t
α = (2.618 - 0.524)/15
α = 0.131 rad/s2
Linear acceleration, a = rα
a = 0.5 × 0.131
a = 0.0655 m/s2
Rounding off to two significant figures, we get,
a = 0.07 m/s2
Therefore, option D is the correct answer.