Question

A puck of mass m1 = 67.0 g and radius r1 = 3.90 cm glides across an air table at a speed of varrowbold = 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 = 125 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 (Figure b).

(B) What is the angular speed about the center of mass?
For (A) I got 6.48e-3 Kgm^2/s

If (a) is correct, how do you find out B? the radius of CoM does not bisect the line of symmetry thus 2 values for r, also is it M*V*R or I*w? In either case, I cannot seem to figure out how to calculate the

Answer 1

M*V*R

Answer 2

What is part a?

Answer 3

My guess is just figure out where the CoM is, then use the distance from it to the little puck's center as the r value




R is the CM vector with the origin at the point of impact, and so the distance to the center of mass of the little puck would be its radius mius the distance to the CM of the system
\[r_{i_{CM}} =r_1-R= .039m+.0255m=.0645m\]
\[M = m_1+m_2\]
Conservation of angular momentum can then be
\[\textbf L_i = \textbf L_f\]
\[\textbf r_{i_{CM}} \times m_1 \textbf v_i = \textbf r_{f_{CM}} \times M \textbf v_f\]
\[r_{i \ _{CM}} m_1 v_i \cancel{\sin \theta}^{\theta=90º \ ; \ =1} = r_{f \ _{CM}}Mv_f \cancel{\sin \varphi}^{\varphi=90º \ ; = \ 1}\]
since the "glue" instantly acts, the r_i is the same for both sides of the equation, just giving
\[ v_f = v_i \frac{m_1}{M}\]

Which makes sense its a completely inelastic collision - all of the little puck's momentum would be given to the two-puck system

Then the angular velocity about the center of mass would be

\[\Large \star \\
\\ \
\\
\omega_{_{CM}} = \frac{v_f}{r_{i_{CM}}}\]



Looking at the full expression makes it seem right
\[ \omega_{_{CM}} = \frac{v_i \frac{m_1}{M}}{r_1-R}\]
\[ = v_i \frac{m_1}{m_1+m_2} \cdot \frac{1}{r_1- \frac{m_1r_1+m_2r_2}{m_1+m_2}}\]
\[ = \frac{v_i m_1}{(m_1+m_2)r_1- (m_1r_1+m_2r_2)}\]
\[=\frac{v_1m_1}{m_2(r_1-r_2)}\]

If the radii stay the same and the little puck's mass gets way bigger, the angular velocity increases - conversely, if the big puck's mass gets bigger, the angular velocity won't be as great.


I think it's right - what do you think??