A puck of mass 80.0 g and radius 3.50 cm slides along an air table at a speed of

Question

A puck of mass 80.0 g and radius 3.50 cm slides along an air table at a speed of 1.50 m/s as shown in the figure below. It makes a glancing collision with a second puck of radius 6.00 cm and mass 120.0 g (initially at rest) such that their rims just touch. Because their rims are coated with instant-acting glue, the pucks stick together and spin after the collision. What is the angular momentum of the system relative to the position of the center of the lab frame? (b) What is the angular speed about the center of mass?

Explanation / Answer

When they touch, the center of mass is distant from the center of the larger puck by
Ycm = m1 ( x1 + x2)/ m1 + m2

= 0.08(0.035+0.06)/0.8+ 0.120 = 0.038 m

Lcm = m1 v1 r1= 0.08(1.5) (0.06) = 7.2 * 10^-3 kg m/s

after collision

Lin = ( I1P w + I2P W)

=( m1 r1^2/2 + m1 d1^2 + m2 r2^2/2 + m2 d2^2) w

7.2 * 10^-3 kg m/s = 0.08( ( 0.035)^2/2 + 0.06^2) + 0.12 (( 0.06)^2/2 + (0.035)^2) w

w= 10.28 rad/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.