Rotational motion - finding tangential acceleration

Click For Summary
The discussion revolves around calculating the tangential and radial components of acceleration for a block released on a frictionless track. The block, with a mass of 6 kg, is initially at a height of 5 m and descends into a half pipe with a radius of 2 m, reaching a height of 2 m. The user successfully calculated the velocity using energy conservation principles but struggles to find the angular acceleration needed for tangential acceleration. Suggestions include considering the forces acting on the block to derive the necessary angular acceleration. The conversation emphasizes the importance of understanding the relationship between linear and angular motion in rotational dynamics.
prettydumbguy
Messages
17
Reaction score
0
1. The problem
A 6 kg block is released from a height of 5 m on a frictionless track and goes into a half pipe with a radius of 2 m. Determine the tangential and radial components of the acceleration when the block reaches a height of 2 m.

Homework Equations


Ac= v^2/r. At = r*angular acceleration. Maybe Vt = angular velocity*r[/B]

The Attempt at a Solution


First I set potential energy at 5m = potential energy at 2m and kinetic energy to find the velocity. I got it and then squared it then divided by 2 to get centripetal/radial acceleration. Now here's where I'm stuck.

I have the radius and the tangential velocity and that's it. I know that tangential acceleration = angular acceleration*r but I don't have the angular acceleration. I can find the instantaneous angular velocity, but it's a single number, not a function so I can't take the derivative. I have no clue where to go from here.
 
Physics news on Phys.org
How about considering the forces acting on the block?
 
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
12
Views
2K
  • · Replies 14 ·
Replies
14
Views
1K
Replies
4
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 9 ·
Replies
9
Views
1K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
15
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 35 ·
2
Replies
35
Views
5K