1 / 9

Dynamics of Collisions and Rotational Motion in Physics

Study the impact dynamics of particles and rods in rotational motion using conservation principles in physics problems.

mkillingsworth
Download Presentation

Dynamics of Collisions and Rotational Motion in Physics

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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

More Related