Solving Plank Pivot Homework: Force Magnitude & Direction

  • Thread starter Thread starter shivam jain
  • Start date Start date
Click For Summary
SUMMARY

The discussion focuses on calculating the force magnitude and direction on a pivoted plank when it is released from a 60-degree angle to the vertical and reaches a horizontal position. The key equations utilized include conservation of energy and the potential energy formula PE = mgl/2cos(θ). The participant attempted to derive the net force using the relationship between gravitational potential energy and rotational kinetic energy but struggled to arrive at the correct answer, indicating a need for clarity on the height (h) used in calculations.

PREREQUISITES
  • Understanding of conservation of energy principles
  • Familiarity with potential energy calculations
  • Basic knowledge of rotational dynamics and moment of inertia
  • Ability to apply trigonometric functions in physics problems
NEXT STEPS
  • Review the concept of gravitational potential energy and its application in pivot systems
  • Study the derivation of moment of inertia for a thin plank
  • Learn about the relationship between angular velocity and linear forces in pivoted systems
  • Explore examples of energy conservation in rotational motion
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and rotational dynamics, as well as educators seeking to clarify concepts related to energy conservation and pivot systems.

shivam jain
Messages
8
Reaction score
0

Homework Statement


a thin plank of mass m and length l is pivoted at 1 end.The plank is released at 60 degrees from the vertical. What is the magnitiude and direction of the force on the pivot when the plank is horizontal.




Homework Equations



conservation of energy

The Attempt at a Solution


when horizontal 2 forces mg and force of pivot. mg we know .force of pivot to find.PE=mgl/2cos theta where theta is angle with vertical.conserving energy i get mgl/2cos theta=1/2*I omega^2.then i found net force m^2 g^2+m*omega^2*l/2 whole underroot.but not getting correct answer
 
Physics news on Phys.org
What makes you think it is not the correct answer?

reason it out - change in gravitational PE is mgh ... what did you use for h and why?
(note: cos(30)=1/2)

Where would the force on the pivot come from?
 

Similar threads

Angular acceleration of plank hanging off edge with a weight
  • · Replies 7 ·
Replies
7
Views
2K
Forces that act on a physical pendulum
  • · Replies 7 ·
Replies
7
Views
3K
Stress on a spaghetti rod balanced from its middle
  • · Replies 6 ·
Replies
6
Views
1K
Simple Harmonic Motion and angular frequency
  • · Replies 1 ·
Replies
1
Views
3K
Equilibrium of Forces and Torques on Sawhorses Supporting a Person on a Board
  • · Replies 6 ·
Replies
6
Views
2K
Force exerted on a pivot point
  • · Replies 2 ·
Replies
2
Views
16K
Angular frequency of plank attached to spring
  • · Replies 1 ·
Replies
1
Views
2K
Tension Physics Help: Solve for T with mg/ma Equations
  • · Replies 4 ·
Replies
4
Views
2K
Rotational Dynamics of a plank of mass
  • · Replies 5 ·
Replies
5
Views
7K
Force on a rod attached to a pivot.
  • · Replies 2 ·
Replies
2
Views
3K