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Problem Set 9

# Problem Set 9 - 8.58 Pendulum Collison The energy of the...

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8.58) Pendulum Collison The energy of the stick can be expressed as a sum of the energy of the effective point mass at the stick center of mass, and the stick’s rotational kinteic energy about that point. The rotational energy is T rot = 1 2 2 = 1 24 ml 2 ˙ θ 2 and the center of mass of the stick is moving along a circle of radius l/ 2, so its kinteic energy will be T trans = 1 2 mv 2 = 1 8 ml 2 ˙ θ 2 adding these together gives us T = 1 6 ml 2 ˙ θ 2 Note that this is the same answer as we would get if we simply use the rotational kinetic energy with moment of inertia about the pivot point. The ball then collides elastically with the stick, meaning if the ball has mass M and its speed after the collision is v , and the stick’s angular speed after the collision is ˙ θ , we must have 1 6 ml 2 ˙ θ 2 = 1 2 Mv 2 + 1 6 ml 2 ˙ θ 2 Using the fact that ˙ θ = ˙ θ/ 2, we see that parenleftbigg 1 6 1 24 parenrightbigg ml 2 ˙ θ 2 = 1 8 ml 2 ˙ θ 2 = 1 2 Mv 2 Now use conservation of angular momentum, using the pivot point as the origin. The angular momentum of the stick before and after the collision can be found from , and the angular momentum of the ball is just Mvl , since its velocity is orthogonal to its displacemenmt from our origin. I ˙ θ = Mvl + I ˙ θ 1 3 ml 2 ˙ θ = Mvl + 1 6 ml 2 ˙ θ 1 6 ml ˙ θ = Mv Substitute this into our previous relation to see 1 8 ml 2 ˙ θ 2 = 1 12 mlv 3 2 l ˙ θ = v and so 1 8 ml 2 ˙ θ 2 = 1 2 M parenleftbigg 3 2 l ˙ θ parenrightbigg 2 m = 9 M We can use conservation of energy to find the stick’s speed when it strikes the ball- the change in potential energy is Δ U = mgl/ 2 since the center of mass has moved from an initial height zero to a new height l/ 2. The kinetic energy must equal this change, so we see 1 6 ml 2 ˙ θ 2 = mgl 2 ˙ θ 2 = 3 g l 1

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Use our expression for v in terms of ˙ θ from conservation of angular momentum to see 3 2 l ˙ θ = 3 2 l radicalbigg 3 g l = v which simplifies to 3 2 radicalBigg 3 gl 2 = v It’s important to note that linear momentum is not conserved- we see ml ˙ θ 2 negationslash = ml ˙ θ 2 + Mv in this case. Their must be an impulse imparted to the pendulum at the pivot point, otherwise the end of the pendulum at the pivot would be moving after the collision. The angular momentum about the pivot must be conserved, since no force applied at the pivot could produce a torque about the pivot. You can see, however, that if the pendulum and the ball were to collide at the pendulum’s center of percussion, both angular and linear momentum would be conserved. 8.68) Center of Percussion We can find the center of percussion by balancing the rotational and translational velocities resulting from the impulse. If the impulse has magnitude P , then the center of mass will be moving with mv CM = P v CM = P m
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