Question

The wad of clay of mass m = 0.36 kg is initially moving with a horizontal velocity v1 = 6.0 m/s when it strikes and sticks to the initially stationary uniform slender bar of mass M = 4.3 kg and length L = 1.64 m. Determine the final angular velocity of the combined body and the x-component of the linear impulse applied to the body by the pivot O during the impact. The angular velocity is positive if counterclockwise, negative if clockwise. The impulse is positive if to the right, negative if to the left.

Answer 1

Complete Question

The diagram for this question is shown on the first uploaded image

Answer:

The angular velocity is
`w_2 = 0.8479 \ rad /s`

The linear impulse applied to the body by the pivot O during the impact is

`O_x \Delta t = 1.2153 N \cdot s`

Step-by-step 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
`m = 0.36 \ kg`

Th velocity is
`v_1 = 6.0 m/s`

The mass of slender bar
`M = 4.3 \ kg`

The length is
`L = 1.64 \ m`

From the diagram the moment of inertia of the slender bar is

`I_o = (1)/(12) ML^2 + M((L)/(6) )^2`

And the moment of inertia of the slender bar and the clay sticked to it is

`I_T = (1)/(9) ML^2 + (1)/(9) mL^2`

`= (1)/(9) [M+m]L^2`

According to the law of conservation of angular momentum

The initial angular momentum = final angular momentum

The initial angular momentum of the clay is

`P_i = mv_1 (L)/(3)`

The final angular momentum is

`P_f = (1)/(9)[M + m] L^2 w_2`

So

`P_ i = P_f \equiv mv_1 (L)/(3) = (1)/(9) (M+ m )L^2 w_2`

So

`w_2 = (3mv_1)/((M+m)L)`

From the diagram the center of gravity is calculated as

`r_G = (M(L)/(6) + m (L)/(3) )/(M+m)`

`= (M + 2m )/(6 (M + m)) L`

Now angular velocity along the x-axis

`-mv_1 + O_x \Delta t = -(M + m )r_G w_2`

`-mv_1 + O_x \Delta t = -(M +m ) (M + 2m)/(6 (M+ m)) L w_2`

`w_2 = (3mv_1)/([M +m]L)`

Where
`O_x \Delta t` is linear impulse applied to the body by the pivot O

Substituting values

`w_2 = (3 * 0.36 * 6)/([4.3 + 0.36] * 1.64)`

`w_2 = 0.8479 \ rad /s`

Making
`O_x \Delta t` the subject of the formula

`O_x \Delta t =(M)/(2( M + m )) mv_1`

substituting value

`O_x \Delta t =(4.3)/(2( 4.2 + 0.36 )) (0.36)* (6.0)`

`O_x \Delta t = 1.2153 N \cdot s`