- #1

Kim Gi Hyuk

- 8

- 0

- Homework Statement
- 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 initial angular velocity is omega_o = 0. At this time, the rod was slightly pushed by an amount d in the longitudinal direction. The acceleration due to gravity is g, and the moment of inertia of the rod about the point of contact is I.

1. When the rod rotates without slipping at the point of contact and leaves the table, find the inclination theta_f and the angular velocity omega_f.

2. If the rod rotates slightly at the point of contact and then slips, find the angle theta_s at which the slip occurs.

3. When leaving situation 2, find the inclination theta_f and the angular velocity omega_f.

- Relevant Equations
- 4. m dv/dt = mg + R + F

where R is normal force and F is frictional force.

5. I d/dt omega = d mg sin theta

where I is rotational inertia moment about contact point.

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 - theta_o )

But I don't know when it leaves the point of contact or what happens if there is slippage.

Ref. European Journal of Physics 16(4):172-176, July 1995.

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 - theta_o )

But I don't know when it leaves the point of contact or what happens if there is slippage.

Ref. European Journal of Physics 16(4):172-176, July 1995.