Does the Average Speed Formula Work for Rolling Objects?

  • Thread starter Thread starter alex_boothby
  • Start date Start date
Click For Summary
SUMMARY

The discussion centers on the application of the average speed formula to rolling objects, specifically a ball bearing rolling down a 5-degree slope. The user calculated translational kinetic energy (0.002520831) and rotational kinetic energy (0.001008332), totaling 0.003529163, which did not match the potential energy (0.014278472) expected from the system. The discrepancy was attributed to the use of average speed (distance/time) instead of instantaneous speed, which is necessary for accurate calculations in accelerating systems.

PREREQUISITES
  • Understanding of kinetic energy concepts: translational and rotational
  • Familiarity with potential energy in gravitational fields
  • Knowledge of angular speed and its calculation using linear velocity and radius
  • Basic principles of motion, particularly for rolling objects
NEXT STEPS
  • Study the differences between average speed and instantaneous speed in physics
  • Learn about energy conservation principles in rolling motion
  • Explore the effects of friction and slipping on rolling objects
  • Investigate the equations of motion for objects rolling down inclined planes
USEFUL FOR

Physics students, educators, and anyone interested in understanding the dynamics of rolling objects and energy conservation principles in mechanics.

alex_boothby
Messages
35
Reaction score
0
when i add ke translational + ke roatational, i don't get potential which i should, although i was wondering if maybe it was due to slipping although i thought it would still be closer than i got

does this sound right?

ke trans(0.002520831)+ke rotational (0.001008332) = 0.003529163
and
pe=0.014278472

as you can see not very near it!

is this right or have i gone wrong?

the exp was rolling a ball bearing down a 5 degree slope and these calcs were from the 1m release mark. the radius is 0.00765m


thanks

to get the angular speed i used linear V/R (radius)

i got linear v by divinding distance by time.
 
Physics news on Phys.org
alex_boothby said:
when i add ke translational + ke roatational, i don't get potential which i should, although i was wondering if maybe it was due to slipping although i thought it would still be closer than i got

does this sound right?

ke trans(0.002520831)+ke rotational (0.001008332) = 0.003529163
and
pe=0.014278472

as you can see not very near it!

is this right or have i gone wrong?

the exp was rolling a ball bearing down a 5 degree slope and these calcs were from the 1m release mark. the radius is 0.00765m


thanks

to get the angular speed i used linear V/R (radius)

i got linear v by divinding distance by time.

If you are rolling a ball down a slope, the linear v will NOT be d/t because the ball is accelerating. d/t is average speed not instantaneous speed, which is what you need for your equations.

-Dan
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 4 ·
Replies
4
Views
1K
Replies
39
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
10
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 7 ·
Replies
7
Views
6K
Replies
14
Views
4K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K