Recent content by Kim Gi Hyuk

Thanks for your comments! I will try to apply the equation for center of mass motion in a rotating coordinate system.

Thanks for your concise and meaningful expanation!

Ok, (1) I confused 'contact point' and 'center of mass'. For the center of mass, I think, is as follows in the rotaional coordinate system, 4-1-1 md omega^2 = mg sin theta - F 4-2-1 md d/dt(omega) = R - mg cos theta (2) If no slip, normal force can become zero only when the rod...

Thank for your comments. v_1 is the velocity along the rod (-F), and v_2 is the velocity in the direction of normal force(R). As you said, both velocities are 0 when maintainig contact. "You only have a linear displacement and, until contact is lost, an angular velocity about the contact...

I tried to solve the problem like this. 4-1. m d/dt (v_1) = mg sin(theta) - F 4-2. m d/dt (v_2) = R - mg cos(theta) If there is no slip, v_1 = v_2 = 0, so F= mg sin(theta), R= mg cos(theta). In the rotational motion equation, we intergrate 5-1. I omega^2 = 2dmg ( sin theta_f -...

Hi! I am a high school physics teacher in Korea. I have a question for you, a physics expert. A rod that falls while rotating from the end of a table A rod of mass m and length L is inclined by theta o (thera_o > 0) when the center of the rod is in contact with the end of the table, and the...