A 1.96 kg toy train rolls around a circular horizontal track. If the train has an angular acceleration of -1.95 rads/s2 and is released with an angular speed of 33.0 rpm, what time is required for the train to come to a complete stop?
See previous post: Sat., 10-22-16, 11:54 AM.
To find the time required for the train to come to a complete stop, we need to use the equations of rotational motion.
First, we need to convert the initial angular speed from rpm to radians per second. There are 2π radians in one revolution, so:
33.0 rpm * (2π radians/1 minute) * (1 minute/60 seconds) = 3.454 radians/second
Now we can use the equation for angular acceleration:
angular acceleration (α) = (change in angular velocity) / (change in time)
In this case, the final angular velocity is 0 (the train comes to a complete stop), the initial angular velocity is 3.454 radians/second, and the angular acceleration is -1.95 radians/second^2.
So we have:
0 = 3.454 radians/second + (-1.95 radians/second^2) * t
Simplifying the equation, we get:
0 = 3.454 - 1.95t
1.95t = 3.454
t = 3.454 / 1.95
t ≈ 1.772 seconds
Therefore, it takes approximately 1.772 seconds for the toy train to come to a complete stop.