Conservation of Angular Momentum and Uniform Rod

Click For Summary
SUMMARY

The discussion centers on a physics problem involving the conservation of angular momentum and a uniform rod. A block of mass 0.157 kg slides down a height of 0.307 m and sticks to a vertical rod of mass 0.397 kg and length 2.78 m, which pivots about point O. The user calculated the moment of inertia (I) as 0.02727 kg*m² and the angular velocity (W) as 4.3357, ultimately finding an angle θ of approximately 40.50 degrees using the provided equations. However, the user expressed uncertainty about the correctness of their solution.

PREREQUISITES
  • Understanding of angular momentum conservation principles
  • Familiarity with moment of inertia calculations
  • Knowledge of kinematic equations related to gravitational potential energy
  • Ability to manipulate trigonometric functions for angle calculations
NEXT STEPS
  • Review the derivation of moment of inertia for composite systems
  • Study the application of conservation of energy in rotational dynamics
  • Explore the use of trigonometric identities in angular motion problems
  • Practice similar problems involving pivoting rods and attached masses
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and rotational dynamics, as well as educators looking for examples of angular momentum applications.

peaceandlove
Messages
66
Reaction score
0

Homework Statement


A small 0.157 kg block slides down a frictionless surface through height h = 0.307 m and then sticks to a uniform vertical rod of mass M = 0.397 kg and length d = 2.78 m. The rod pivots about point O through angle θ before momentarily stopping. Find θ.


Homework Equations


Equation 1: I= (1/3)Md^2 + md^2
Equation 2: v=(2gh)^(1/2)
Equation 3: W=m(block)dv/I
Equation 4: arccos(1 -(I*W^2/[d*g(2m + M) ]))

[I and W are for the road and block system]


The Attempt at a Solution


Using equation 1, I found I to be 0.02727 kg*m^2. Then I found W to be 4.3357 using equation 2 and 3. I then plugged in the values into equation 4 and got 40.50412 degrees, but apparently I'm doing something wrong.
 
Physics news on Phys.org
Nevermind!
 

Similar threads

Replies
10
Views
3K
Replies
17
Views
2K
  • · Replies 17 ·
Replies
17
Views
1K
  • · Replies 71 ·
3
Replies
71
Views
5K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 8 ·
Replies
8
Views
5K
  • · Replies 46 ·
2
Replies
46
Views
5K
Replies
18
Views
7K
  • · Replies 25 ·
Replies
25
Views
4K
Replies
21
Views
3K