Minimum Speed for Balsa Cube Tipping Problem

Click For Summary
The discussion revolves around calculating the minimum speed required for a bullet to tip a balsa cube when it embeds into the cube. A participant initially calculated a speed of 921 m/s but expressed uncertainty about its accuracy. Another participant criticized the approach, emphasizing the need for a deeper understanding of the physics involved rather than just plugging values into an equation. They suggested drawing a diagram and applying general principles learned in class. The conversation highlights the importance of conceptual understanding in solving physics problems.
T$$$
Messages
2
Reaction score
0
A solid balsa cube of side length L = 16.0” and mass M = 8.60 kg is at rest on a horizontal table top. It is constrained to rotate about a fixed and frictionless axis, AB, along one edge of the cube. A bullet of mass m = 50.0 g is fired with speed v at the other side of the cube, at height a = 12.0” above the table surface. The bullet becomes embedded in the cube in the middle of the face opposite face ABCD. Find the minimum value of the speed v required to tip the cube over, so that it falls on face ABCD. You may assume that the bullet mass m is small enough, compared to M, that it does not change the rotational inertia or center of mass of the cube after it embeds.

I've been working on this problem for a while and the only answer i got was 921 m/s but i don't think that's right... if someone could help me set up the problem that would be appreciated.
 
Physics news on Phys.org
How did you get to 921 m/s? Then we can see where you went wrong.
 
Greg Bernhardt said:
How did you get to 921 m/s? Then we can see where you went wrong.
I honestly have no idea how to set it up, i just plugged it into an equation my professor showed me on a similar problem today. which is mvr=.5ml^2 +md^2
so .05*v*4=1/12 *8.6 *16^2 +.25*4^2
 
T$$$ said:
I just plugged it into an equation my professor showed me on a similar problem today.
That sounds like a recipe for failure. There is too much specific detail here for such an approach. There's no substitute for understanding and applying the general principles.
I gather that AB is an edge on the table. Have you drawn a diagram? What general principles have you learned that can be applied?
 
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 20 ·
Replies
20
Views
2K
Replies
7
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 7 ·
Replies
7
Views
5K
Replies
30
Views
3K
  • · Replies 23 ·
Replies
23
Views
16K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 6 ·
Replies
6
Views
7K