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Motion in a Circle Question

  • Thread starter dman_PL
  • Start date
  • #1
15
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Homework Statement


Can someone check if this is right? The time seems okay, but the work I feel is wrong

A ball with moment of inertia 0.1kg · m2 is rotating on a table, but
friction is slowing it down with constant angular acceleration. The ball is originally spinning at 2π radians per second, but slows down to 10 percent of that value in 2 seconds


1) How long does it take the ball to come to rest (time since it began spinning)
2)How much work is done on the ball to bring it to rest


Homework Equations


1)Since angular acceleration is constant, we use ∆ω = α∆t
2)W = ∆K = Kf − Ki = 0 −1/2Iω


The Attempt at a Solution


1) Since I have the ball spinning at 2π/sec, and it slows down -10% of the original speed in two seconds, so i plug in the information into the equation. so Δw= (-.10)(2π)/2sec which equals -.314159. Which then I can solve for t. New equation: t=Δw/α. So -.314159/-.628318= 5 Sec.
2) (-1/2)(.1kg*m^2)(.628318)^2=-.0197J
 

Answers and Replies

  • #2
Andrew Mason
Science Advisor
Homework Helper
7,567
330

Homework Statement


Can someone check if this is right? The time seems okay, but the work I feel is wrong

A ball with moment of inertia 0.1kg · m2 is rotating on a table, but
friction is slowing it down with constant angular acceleration. The ball is originally spinning at 2π radians per second, but slows down to 10 percent of that value in 2 seconds


1) How long does it take the ball to come to rest (time since it began spinning)
2)How much work is done on the ball to bring it to rest


Homework Equations


1)Since angular acceleration is constant, we use ∆ω = α∆t
2)W = ∆K = Kf − Ki = 0 −1/2Iω
Check this. The rotational kinetic energy has dimensions of kg m^2/sec^2. This has dimension of kg m^2/sec.

The Attempt at a Solution


1) Since I have the ball spinning at 2π/sec, and it slows down -10% of the original speed in two seconds, so i plug in the information into the equation. so Δw= (-.10)(2π)/2sec which equals -.314159. Which then I can solve for t. New equation: t=Δw/α. So -.314159/-.628318= 5 Sec.
2) (-1/2)(.1kg*m^2)(.628318)^2=-.0197J
Better to find the algebraic solution and then plug in numbers.

First, find the angular deceleration α. Then use that to find the time it takes to stop using the correct expression for rotational energy.

AM
 
  • #3
15
0
Check this. The rotational kinetic energy has dimensions of kg m^2/sec^2. This has dimension of kg m^2/sec.
Better to find the algebraic solution and then plug in numbers.

First, find the angular deceleration α. Then use that to find the time it takes to stop using the correct expression for rotational energy.

AM
So to find α, I am still using Δω/Δt correct?
 
  • #4
Andrew Mason
Science Advisor
Homework Helper
7,567
330
So to find α, I am still using Δω/Δt correct?
Correct. You don't have to use energy to work out the time it takes to stop. Just use α = Δω/Δt. What is the change in ω when it stops? Knowing α and Δω you can work out the Δt.

AM
 

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