A 1.11 kg toy train rolls around a circular horizontal track. If the train has an angular acceleration of -2.2 rads/s2 and is released with an angular speed of 17.0 rpm, what time is required for the train to come to a complete stop?
Vo = 17rev/min. * 6.28rad/rev * 1min/60s = 1.78 rad/s.
V = Vo + a*t.
0 = 1.78 - 2.2*t, t = ?.
To solve this problem, we need to first convert the given angular acceleration and angular speed into the same unit.
Given:
Mass of the toy train, m = 1.11 kg
Angular acceleration, α = -2.2 rad/s^2 (negative sign indicates deceleration)
Initial angular speed, ω₀ = 17.0 rpm
First, we need to convert the initial angular speed from rpm to rad/s.
1 revolution = 2π radians
1 minute = 60 seconds
So,
ω₀ = 17.0 rpm * (2π rad/1 revolution) * (1 minute/60 seconds)
ω₀ = 17.0 * (2π/60) rad/s
Now, we can use the kinematic equation of rotational motion to find the time (t) required for the train to stop:
ω = ω₀ + αt
Since we are trying to find the time when the train comes to a complete stop, we can substitute ω = 0:
0 = ω₀ + αt
Solving for t:
αt = -ω₀
t = (-ω₀) / α
Substituting the given values:
t = (-17.0 * (2π/60) rad/s) / -2.2 rad/s^2
Now, we can calculate the value of time (t) to find the answer.