Ekpyrotic Posted December 7, 2008 Posted December 7, 2008 I have a rod with a fixed base, a hinge. It starts to fall sideways (rotating). Using the conservation of energy, how is the angular speed related to the angle with the ground?
D H Posted December 7, 2008 Posted December 7, 2008 This looks a lot like homework, and we have a policy here of not doing your homework for you. How would you describe this problem mathematically? What are the relevant equations?
Ekpyrotic Posted December 7, 2008 Author Posted December 7, 2008 (edited) Here's where I am. The object has grav. potential at the start, then transfers that to kinetic energy. [math]gpe = \frac{1}{2} I \omega^2 + \frac{1}{2}mv_x^2 + \frac{1}{2}mv_y^2[/math] We can work out [math]I[/math], which is equal to [math]\frac{1}{3} mL^2[/math] Any advice? Hints? Edited December 7, 2008 by Ekpyrotic multiple post merged
D H Posted December 7, 2008 Posted December 7, 2008 There are several ways to look at this problem: 1. As a combination of rotational and translational motion. In general, the motion of any object can be expressed as a combination of rotation about the center of mass plus translation of the center of mass. You haven't done that in this case. 2. As a purely rotational problem. If you choose the right representation you won't have any translational energy to worry about. 3. As a pure translation problem. The expression [math]1/2 I \omega^2[/math] is derived by looking at the contribution to kinetic energy by each infinitesimal mass element in a solid object. ======================= There is no reason to separate the velocity of some point on the rod into x and y components. All you care about is the square of the magnitude of the velocity vector. What is the speed at which some point on the rod is moving given that the rod has angular velocity [math]/omega[/math]?
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now