Skip to main content
Ch 09: Rotation of Rigid Bodies

Chapter 9, Problem 9

A high-speed flywheel in a motor is spinning at 500 rpm when a power failure suddenly occurs. The flywheel has mass 40.0 kg and diameter 75.0 cm. The power is off for 30.0 s, and during this time the flywheel slows due to friction in its axle bearings. During the time the power is off, the flywheel makes 200 complete revolutions. (a) At what rate is the flywheel spinning when the power comes back on?

Verified Solution
Video duration:
5m
This video solution was recommended by our tutors as helpful for the problem above.
1265
views
Was this helpful?

Video transcript

Welcome back everybody. We are given some sort of rotational shaft here and we are told that as it is spinning, it can only a maximum speed of 450 rpm. And we're gonna say at time zero, right, it reaches this maximum and it powers down. So our initial speed here is going to be 450 revolution per minute. Now, after it makes a total number 140 revolutions, which happens happens in a time period of 23 seconds, it's reconnected and it starts to spin again. But we want to figure out when it's reconnected. What is the speed then, after it has slowed down, we're actually gonna use a cinematic formula for angular motion here. That states that our change in angular position is equal to 1/2 times our initial angular velocity plus our final angular velocity I'm T. And we can go ahead and plug in the values that we know. But before doing that I'm actually going to isolate what we want to find which is this guy right here, let me multiply both sides by two over T and then subtract our initial angular velocity. We get that our final angular velocity equal to two times are changin angular position over our time period minus our initial angular velocity. Now let's go ahead and plug in some numbers we have that are final angular velocity equal to two times 1 40/23 minus 75 giving us 4. revolutions per one second. But we want our units to be revolutions per minute. So we're gonna multiply this by 60 seconds, since there are 60 seconds in one minute, giving us a final answer of 280 r e. M, corresponding to answer choice B. Thank you all so much for watching. Hope this video helped. We will see you all in the next one.
Related Practice
Textbook Question
A wheel is rotating about an axis that is in the z-direction. The angular velocity ω_z is -6.00 rad/s at t = 0, increases linearly with time, and is +4.00 rad/s at t = 7.00 s. We have taken counterclockwise rotation to be positive. (c) What is the angular displacement of the wheel at t = 7.00 s?
692
views
Textbook Question
An electric fan is turned off, and its angular velocity decreases uniformly from 500 rev/min to 200 rev/min in 4.00 s. (a) Find the angular acceleration in rev/s^2 and the number of revolutions made by the motor in the 4.00-s interval. (b) How many more seconds are required for the fan to come to rest if the angular acceleration remains constant at the value calculated in part (a)?
1001
views
Textbook Question
An electric fan is turned off, and its angular velocity decreases uniformly from 500 rev/min to 200 rev/min in 4.00 s. (b) How many more seconds are required for the fan to come to rest if the angular acceleration remains constant at the value calculated in part (a)?
1855
views
Textbook Question
CALC The angle θ through which a disk drive turns is given by θ(t) = a + bt - ct^3, where a, b, and c are constants, t is in seconds, and θ is in radians. When t = 0, θ = p/4 rad and the angular velocity is 2.00 rad/s. When t = 1.50 s, the angular acceleration is 1.25 rad/s^2. (a) Find a, b, and c, including their units.
596
views
Textbook Question
CALC The angle θ through which a disk drive turns is given by θ(t) = a + bt - ct^3, where a, b, and c are constants, t is in seconds, and θ is in radians. When t = 0, θ = p/4 rad and the angular velocity is 2.00 rad/s. When t = 1.50 s, the angular acceleration is 1.25 rad/s^2. (b) What is the angular acceleration when θ = p/4 rad?
1613
views
Textbook Question
CALC The angle θ through which a disk drive turns is given by θ(t) = a + bt - ct^3, where a, b, and c are constants, t is in seconds, and θ is in radians. When t = 0, θ = p/4 rad and the angular velocity is 2.00 rad/s. When t = 1.50 s, the angular acceleration is 1.25 rad/s^2. (c) What are θ and the angular velocity when the angular acceleration is 3.50 rad/s^2?
876
views