Angular Momentum: Puck Spinning

  • #1
95
2

Homework Statement


The puck in the figure shown below has a mass of 0.120 kg. Its original distance from the center of rotation is 40.0 cm, and it moves with a speed of 80.0 cm/s. The string is pulled downward 15.0 cm through the hole in the frictionless table. Determine the work done on the puck.

4gWOb2a.png

Homework Equations



## L = I \omega##
##KE = 1/2 \cdot I \cdot \omega^2 ##

The Attempt at a Solution



Consider angular momentum. Note it's conserved.

Therefore, ## \frac{1}{2}mr_i^2 \omega_i = \frac{1}{2} mr_f^2 \omega_f \Rightarrow r_i^2 \omega_i =r_f^2 \omega_f ##

Note that ##\omega_i = 80 cm/s = 2 rad/s## (because 2pi radians is one revolution, which is 80pi cm, so 1 radian is 40 cm)

Threrefore, ##0.40^2 \cdot 2 = 0.25^2 \cdot \omega_f \Rightarrow \omega_f = 5.12##

Note that Work Done is the change in kinetic energy. KE initial is equal to ## 1/2 (1/2 \cdot 0.12 \cdot 0.4*0.4) \cdot 2^2 ## (1/2 Iw^2 = 1/2 (1/2mr^2)w^2). KE final is simialrly equal to ## 1/2 (1/2 \cdot 0.12 \cdot 0.25 \cdot 0.25 \cdot 5.12^2##.

We can do the calculation, to get the change as 0.029952. But the book says answer is 5.99*10^-2.

What's wrong?
 

Answers and Replies

  • #2
Nathanael
Homework Helper
1,650
239
KE initial is equal to ## 1/2 (1/2 \cdot 0.12 \cdot 0.4*0.4) \cdot 2^2 ## (1/2 Iw^2 = 1/2 (1/2mr^2)w^2). KE final is simialrly equal to ## 1/2 (1/2 \cdot 0.12 \cdot 0.25 \cdot 0.25 \cdot 5.12^2##.
What's the (extra) factor of 1/2 for? (I see an extra one in both initial and final KE's)

You don't even need to bother with angular velocities. (You can solve it that way if you'd like, but it's redundant.) You can just use the angular momentum of a particle = mvr
 
  • #3
tms
644
17
##KE = 1/2 \cdot I \cdot \omega^2 ##

##\ldots##

Therefore, ## \frac{1}{2}mr_i^2 \omega_i = \frac{1}{2} mr_f^2 \omega_f \Rightarrow r_i^2 \omega_i =r_f^2 \omega_f ##
Do you see the discrepancy between the two quoted lines?
 
Last edited:
  • #4
95
2
What's the (extra) factor of 1/2 for? (I see an extra one in both initial and final KE's)

You don't even need to bother with angular velocities. (You can solve it that way if you'd like, but it's redundant.) You can just use the angular momentum of a particle = mvr
How do you get that angular momentum of a particle = mvr

@tms, EDIT: No, the kinetic energy isn't being calculated there. That's the conservation of angular momentum, which is Iw.

FINAL EDIT: I see the problem now. Moment of inertia is mr^2, not 1/2 mr^2.
 
  • #5
Nathanael
Homework Helper
1,650
239
How do you get that angular momentum of a particle = mvr
Angular momentum of a particle equals Iω, but I=mr2, so the angular momentum is mωr2
Since, ωr=v, the angular velocity is mvr
 
  • #6
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
35,577
6,449
Angular momentum of a particle equals Iω, but I=mr2, so the angular momentum is mωr2
Since, ωr=v, the angular velocity is mvr
You mean, the angular momentum is mvr, yes?
 
  • #7
Nathanael
Homework Helper
1,650
239
You mean, the angular momentum is mvr, yes?
Oops! Mindless typo, thanks.
 

Related Threads on Angular Momentum: Puck Spinning

Replies
1
Views
3K
  • Last Post
Replies
0
Views
4K
Replies
8
Views
12K
Replies
3
Views
759
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
2
Views
3K
  • Last Post
2
Replies
35
Views
2K
  • Last Post
Replies
1
Views
3K
Replies
1
Views
11K
Top