Angular energy vs translational energy

  • Thread starter GnG.Vike13
  • Start date
  • #1
11
0
i don't understand this at all.... if we tip over a rectangular block (such that a corner is its pivot), will it have rotational kinetical energy, translational kinetic energy, or both when it hits the ground???
 

Answers and Replies

  • #2
sophiecentaur
Science Advisor
Gold Member
2020 Award
25,514
5,025
Ask yourself "Is its centre of mass moving horizontally and or vertically with respect to the floor when it actually lands and is it rotating?"
How would things be different if there were no friction between the corner and the table (assuming you placed it just beyond its tipping angle?
What does that tell you about its translational and rotational KE, bearing in mind that the same amount of energy is available in each case.
 
Last edited:
  • #3
11
0
so the translational kinetic energy is based on its center of mass' velocity and the rotational energy is based on the I of the whole block and w?

and, just to be clear, the weight vector will always come from the center of mass, correct?

also, how would friction play into this? would it be at the axis of rotation, thus causing not torque, but preventing it from making a translational acceleration?
 
  • #4
sophiecentaur
Science Advisor
Gold Member
2020 Award
25,514
5,025
so the translational kinetic energy is based on its center of mass' velocity and the rotational energy is based on the I of the whole block and w?

and, just to be clear, the weight vector will always come from the center of mass, correct?

also, how would friction play into this? would it be at the axis of rotation, thus causing not torque, but preventing it from making a translational acceleration?
That seems the right idea. But there would be a definite difference if there were no friction. What would the block rotate around if the corner were allowed to slip? Would the cm ever be moving horizontally at all? Would the block take the same time to fall with and without friction?
Your statements about mvsquared/2 and Iωsquared/2 are obviously correct. I think the only way that there would be no translational energy would be if it were just a couple that was applied to the block so that the cm was not moving in any direction. This is definitely not the case here.
 
  • #5
11
0
That seems the right idea. But there would be a definite difference if there were no friction. What would the block rotate around if the corner were allowed to slip? Would the cm ever be moving horizontally at all? Would the block take the same time to fall with and without friction?
Your statements about mvsquared/2 and Iωsquared/2 are obviously correct. I think the only way that there would be no translational energy would be if it were just a couple that was applied to the block so that the cm was not moving in any direction. This is definitely not the case here.
okay, it's good to know that i understand this. one more question, perhaps.

what happens if the block isn't tipping over and the frictional force is applied completely on block's bottom side? how do you look at the problem then?
 
  • #6
sophiecentaur
Science Advisor
Gold Member
2020 Award
25,514
5,025
Then, with sufficient force applied, won't it just slide with some translational KE and no rotational energy?
 

Related Threads on Angular energy vs translational energy

Replies
3
Views
18K
Replies
1
Views
5K
  • Last Post
Replies
4
Views
2K
Replies
2
Views
1K
  • Last Post
Replies
6
Views
795
  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
7
Views
12K
  • Last Post
Replies
2
Views
5K
  • Last Post
Replies
2
Views
10K
  • Last Post
Replies
2
Views
1K
Top