A small puck of mass 20 g and radius 8 cm slides along an air table with a speed
ID: 1913859 • Letter: A
Question
A small puck of mass 20 g and radius 8 cm slides along an air table with a speed of 1.2 m/s. It makes a glazing collision with a larger puck of radius 28 cm and mass 72 g (initially at rest) such that their rims just touch. Both are disks with a moment of A small puck of mass 20 g and radius 8 cm slides along an air table with a speed of 1.2 m/s. It makes a glazing collision with a larger puck of radius 28 cm and mass 72 g (initially at rest) such that their rims just touch. Both are disks with a moment of inertia equal to1/2m r^2. The pucks stick together and spin around after the collision. After the collision the center of mass CM has a linear velocity v and an angular velocity ? about the CM.
1. What is the angular momentum of the system relative to the center of mass after the collision? Answer in units of kg ? m2/s
2. What is the system?s angular speed about the CM after the collision? Answer in units of rad/s
A small puck of mass 20 g and radius 8 cm slides along an air table with a speed of 1.2 m/s. It makes a glazing collision with a larger puck of radius 28 cm and mass 72 g (initially at rest) such that their rims just touch. Both are disks with a moment of A small puck of mass 20 g and radius 8 cm slides along an air table with a speed of 1.2 m/s. It makes a glazing collision with a larger puck of radius 28 cm and mass 72 g (initially at rest) such that their rims just touch. Both are disks with a moment of inertia equal to1/2m r^2. The pucks stick together and spin around after the collision. After the collision the center of mass CM has a linear velocity v and an angular velocity ? about the CM. 1. What is the angular momentum of the system relative to the center of mass after the collision? Answer in units of kg ? m2/s 2. What is the system?s angular speed about the CM after the collision? Answer in units of rad/sExplanation / Answer
Place the origin of the coordinate system at the center of puck 2, with the pucks at the moment of contact having their centers on the y-axis. The location of the center of mass from the origin is ym = m1/(m1+m2) * (r1+r2) = cm The angular momentum of the system the instant before contact, measure from the system's center of mass is m1*v1*(distance from center of m1 to center of mass)= m1*v1*(r1 + r2 - ym) The angular speed equals the linear speed divided by the radius from the center or rotation, which is the center of mass in this case. So after the collision: m1*Va*r1^2 + m2*Va*r2^2
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