h = ?

1. v w v

2. v h = ?

3. M d The particle of mass m in the diagram slides down the frictionless surface through height h and collides with the uniform vertical rod (of mass M and length d). sticking to it. The rod pivots about point 0 through the angle q before momentarily stopping. Find speed of m at bottom of ramp Find the angular velocity after the collision Find q

4. M d Speed at the bottom of the ramp Angular velocity after the collision a) Conservation of Energy b) Conservation of Angular Momentum c) Angular velocity

5. M = 1.0 kg 0.60 cm m = 0.20 kg A uniform rod (length = 0.60 m, mass = 1.0 kg) rotates about an axis through one end. As the rod swings through its lowest position, the end of the rod collides with a small 0.20 kg putty wad that sticks to the end of the rod.

6. M = 1.0 kg 0.60 cm 2.4 rad/s m = 0.20 kg If the angular speed of the rod just before the collision is 2.4 rad/s, what is the angular speed of the rod-putty system immediately after the collision? Conservation of Angular Momentum w = 1.5 rad/s

7. Ballistic Pendulum I(rod) = 0.060 kgm2 A 1.0 g bullet is fired into a 0.50 kg block that is mounted on the end of a 0.60 m non-uniform rod of mass 0.50 kg. The block-rod-bullet system then rotates about a fixed axis. The rotational inertia of the rod alone about A is 0.060 kg m2. Assume the block is small enough to treat as a particle on the end of the rod 0.60 m v m = 1.0 g M = 0.50 kg

8. A (a) What is the rotational inertia of the block-rod-bullet system about point A? Irod = 0.060 kg m2 r = 0.60 m I = Irod + Smr2 I = Irod + (m + M)r2 I = 0.060 + (0.501)(0.60)2 m = 1.0 g M = 0.50 kg I = 0.24 kg m2

9. (b) If the angular speed of the system about A just after the bullet's impact is 4.5 rad/s, what is the speed of the bullet just before the impact? A Conservation of Angular Momentum r = 0.60 m m = 1.0 g v w = 4.5 rad/s I = 0.24 kg m2 v = 180 m/s