A puck of mass m1 = 63.0 g and radius r1 = 4.10 cm glides across an air table at

Question

A puck of mass m1 = 63.0 g and radius r1 = 4.10 cm glides across an air table at a speed of = 1.50 m/s as shown in Figure a. It makes a glancing collision with a second puck of radius r2 = 6.00 cm and mass m2 = 115 g (initially at rest) such that their rims just touch. Because their rims are coated with instant-acting glue, the pucks stick together and rotate after the collision.

(a) What is the magnitude of the angular momentum of the system relative to the center of mass?

(b) What is the angular speed about the center of mass?

Explanation / Answer

mass of first puck m1 =63g

radius of first puck r1 =4 cm =4*10^-2 m

speed of first puck v1 =1.5 m/s

mass of second puck m2 =115g

radius of second puck r2 =6 cm =6*10^-2 m

speed of second puck v2 =0 m/s

a)

angular momentum of the system relative to the centre of mass is

L =m1v1r1+m2v2r2

L =(63*10^-3 kg)(1.5 m/s)(4*10^-2 m)+0

L =3.78*10^-3 kg.m^2/s

b)

moment of inertia about the centre of mass

law of conservation of angular momentum,

L =I

therefore, angular speed =L/I

=4.89 rad/s

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