Rotational Inertia of Pulley, Integration Help

Click For Summary
SUMMARY

The discussion focuses on calculating the angular acceleration and angular velocity of a pulley with a rotational inertia of 1.5 x 10-3 kg·m2 and a radius of 21 cm, subjected to a time-varying tangential force described by F(t) = 0.50t + 0.30t2. At t = 10.0 seconds, the angular acceleration is determined to be 4900 rad/s2. To find the angular velocity at the same time, integration of the angular acceleration function is required, leading to the formula ω|10 = 140 * ∫(0.50t + 0.30t2) dt evaluated from 0 to 10.

PREREQUISITES
  • Understanding of rotational dynamics, specifically torque and angular acceleration.
  • Familiarity with the concept of rotational inertia (moment of inertia).
  • Basic knowledge of calculus, particularly integration techniques.
  • Ability to apply Newton's second law in rotational motion contexts.
NEXT STEPS
  • Learn integration techniques for functions involving polynomials, specifically ∫(at + bt2) dt.
  • Study the relationship between torque, angular acceleration, and moment of inertia using the equation Tnet = Iα.
  • Explore the concept of angular velocity and how it relates to angular acceleration over time.
  • Investigate the effects of varying forces on rotational motion and how to model them mathematically.
USEFUL FOR

Students studying physics, particularly those focusing on rotational dynamics, as well as educators seeking to explain the integration of forces in motion. This discussion is also beneficial for anyone needing to understand the application of calculus in physical systems.

rmunoz
Messages
29
Reaction score
0

Homework Statement



A pulley, with a rotational inertia of 1.5 10-3 kg·m2 about its axle and a radius of 21 cm, is acted on by a force applied tangentially at its rim. The force magnitude varies in time as F = 0.50t + 0.30t^{2}, where F is in Newtons and t in seconds. The pulley is initially at rest.

(a) At t = 10.0 s what is its angular acceleration?
rad/s2


(b) At t = 10.0 s what is its angular velocity?
rad/s




Homework Equations


Tnet=I\alpha

f=.50t + .30t^{2}

The Attempt at a Solution


I already got the acceleration for the pulley. It turned out to be 4900 rad/sec. But now i assume because the amount of force is reliant on the time (in other words, is accelerating at a non-constant rate), some integration will be needed. The problem for me is, i have only the vaguest idea of how to actually integrate. Would anyone mind helping me with this problem if it does in fact require integration, by showing me the exact steps? And if that is not the case, perhaps point me in the right direction?
 

Attachments

  • screen-capture-4.png
    screen-capture-4.png
    18.1 KB · Views: 618
Physics news on Phys.org
rmunoz said:

Homework Statement



A pulley, with a rotational inertia of 1.5 10-3 kg·m2 about its axle and a radius of 21 cm, is acted on by a force applied tangentially at its rim. The force magnitude varies in time as F = 0.50t + 0.30t^{2}, where F is in Newtons and t in seconds. The pulley is initially at rest.

(a) At t = 10.0 s what is its angular acceleration?
rad/s2

(b) At t = 10.0 s what is its angular velocity?
rad/s


Homework Equations


Tnet=I\alpha

f=.50t + .30t^{2}

The Attempt at a Solution


I already got the acceleration for the pulley. It turned out to be 4900 rad/sec. But now i assume because the amount of force is reliant on the time (in other words, is accelerating at a non-constant rate), some integration will be needed. The problem for me is, i have only the vaguest idea of how to actually integrate. Would anyone mind helping me with this problem if it does in fact require integration, by showing me the exact steps? And if that is not the case, perhaps point me in the right direction?

You are correct. You will need to do an integral to determine the angular velocity.

You have already evaluated the a(t) at 10 sec to determine a|10 from

|a(t)| = |r|*|F(t)|/I = 140*|.5*t + .3*t2|

Since a(t) = dω/dt

then ω|10 = ∫a(t)*dt

or

ω|10 = 140*∫(.5t+.3t2)*dt = 140*(.25*t2 +.1*t3 + c) evaluated from 0 to 10.

Since you are told it was at rest at t=0, then the constant of integration c = 0.
 
Awesome, thank you for the walkthrough, that was incredibly helpful
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 5 ·
Replies
5
Views
3K
Replies
8
Views
14K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 30 ·
2
Replies
30
Views
4K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
Replies
2
Views
8K
Replies
9
Views
2K