Complete Question
The diagram for this question is shown on the first uploaded image
Answer:
The angular velocity is [tex]w_2 = 0.8479 \ rad /s[/tex]
The linear impulse applied to the body by the pivot O during the impact is
[tex]O_x \Delta t = 1.2153 N \cdot s[/tex]
Explanation:
The free body diagram of the image is shown on the second uploaded image
From the question we are told that
The mass of the wad of clay is [tex]m = 0.36 \ kg[/tex]
Th velocity is [tex]v_1 = 6.0 m/s[/tex]
The mass of slender bar [tex]M = 4.3 \ kg[/tex]
The length is [tex]L = 1.64 \ m[/tex]
From the diagram the moment of inertia of the slender bar is
[tex]I_o = \frac{1}{12} ML^2 + M(\frac{L}{6} )^2[/tex]
And the moment of inertia of the slender bar and the clay sticked to it is
[tex]I_T = \frac{1}{9} ML^2 + \frac{1}{9} mL^2[/tex]
[tex]= \frac{1}{9} [M+m]L^2[/tex]
According to the law of conservation of angular momentum
The initial angular momentum = final angular momentum
The initial angular momentum of the clay is
[tex]P_i = mv_1 \frac{L}{3}[/tex]
The final angular momentum is
[tex]P_f = \frac{1}{9}[M + m] L^2 w_2[/tex]
So
[tex]P_ i = P_f \equiv mv_1 \frac{L}{3} = \frac{1}{9} (M+ m )L^2 w_2[/tex]
So
[tex]w_2 = \frac{3mv_1}{(M+m)L}[/tex]
From the diagram the center of gravity is calculated as
[tex]r_G = \frac{M\frac{L}{6} + m \frac{L}{3} }{M+m}[/tex]
[tex]= \frac{M + 2m }{6 (M + m)} L[/tex]
Now angular velocity along the x-axis
[tex]-mv_1 + O_x \Delta t = -(M + m )r_G w_2[/tex]
[tex]-mv_1 + O_x \Delta t = -(M +m ) \frac{M + 2m}{6 (M+ m)} L w_2[/tex]
[tex]w_2 = \frac{3mv_1}{[M +m]L}[/tex]
Where [tex]O_x \Delta t[/tex] is linear impulse applied to the body by the pivot O
Substituting values
[tex]w_2 = \frac{3 * 0.36 * 6}{[4.3 + 0.36] * 1.64}[/tex]
[tex]w_2 = 0.8479 \ rad /s[/tex]
Making [tex]O_x \Delta t[/tex] the subject of the formula
[tex]O_x \Delta t =\frac{M}{2( M + m )} mv_1[/tex]
substituting value
[tex]O_x \Delta t =\frac{4.3}{2( 4.2 + 0.36 )} (0.36)* (6.0)[/tex]
[tex]O_x \Delta t = 1.2153 N \cdot s[/tex]
