Answer to Question #119115 in Classical Mechanics for Stan

Question

A puck of mass 80.0 g and radius 4.00 cm slides along an
air table at a speed of 1.50 m/s. It makes a glancing collision with a second puck
of radius 6.00 cm and mass 120 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 (Fig. P11.34b). (a) What is the angular
momentum of the system relative to the center of mass?
(b) What is the angular speed about the center of mass?

Answer 1

As per the given question,

mass of the puck $(m_1)=80.0 gm$

radius $(r_1)=4.0cm$

Speed of the puck $(v_1)= 1.50 m/s$

Radius of the glancing collision $(r_2)=6.00 cm$

mass $(m_2)=120 gm$

As the diagram is not given in the question,

Applying the conservation of center of mass,

$y=\dfrac{m_1r_1+m_2r_2}{m_1+m_2}$

now substituting the values,

$y=\dfrac{80\times4-120\times 6}{80+120}=-2cm$

Hence angular momentum of the $(L)=mvr =80\times10^{-3} \times2\times 10^{-2}\times 1.5 kg m^2/sec$

$=7.2\times 10^{-3}kg m^2/sec$

b) angular speed about the center of mass $(\omega)=\dfrac{L}{I}$

$=\dfrac{7.2\times 10^{-3}kg m^2/sec}{I}$

$I=\frac{m_1 r_1^2}{2}+m_1d_1^2+\frac{m_2 r_2^2}{2}+m_2d_2^2$

$=\dfrac{80\times 10^{-3}\times (4\times 10^{-2})^2}{2}+80\times 10^{-3}\times (6\times 10^{-2})^2+\dfrac{120\times 10^{-3}\times (6\times 10^{-2})^2}{2}+{120\times 10^{-3}\times (6\times 10^{-2})^2}$ $I=7.6\times 10^{-4}kg-m^2$

Now, substituting the values,

$(\omega)=\dfrac{7.2\times 10^{-3}kg-m^2/sec}{7.6\times 10^{-4}kg-m^2}$ $=9.47 rad/sec$