Respuesta :
Answer:
a) Before:
I = 0.00928 kg*m^2
after:
I = 0.00949 kg*m^2
b) W = 65.25 rad/s
c) k = 20.2021 J
Explanation:
a)The moment of inertia before the collision is done by:
I = [tex]\frac{1}{3}ML^2[/tex]
where M is the mass and L is the length.
I = [tex]\frac{1}{3}(0.116 kg)(0.49 m)^2[/tex]
I = 0.00928 kg*m^2
After the collision the moment of inertia is:
I = [tex]\frac{1}{3}ML^2 + M_mR^2[/tex]
the first part is the moment of inertia before the collision but now we plus the mass of the putty ball ([tex]M_m[/tex] and the distance R where the putty ball strickes.
I = [tex]\frac{1}{3}(0.116)(0.49)^2 + (0.014)(0.49/4)^2[/tex]
I = 0.00949 kg*m^2
b)
For answer this we will use the law of the conservation of the angular momentum:
[tex]L_i = L_f[/tex]
Adittionally, the angular momentum L could be calculated by:
L = MDV
where M is the mass, V is the velocity and D is the lever arm, or:
L = IW
where I is the moment of inertia and W is the angular velocity.
Following the equations, now we have:
[tex]M_pV_pD = I_sW[/tex]
where [tex]M_p[/tex] is the mass of the putty ball, [tex]V_p[/tex] is the velocity of the putty ball, and [tex]I_s[/tex] is the moment of inertia after the collition
replacing the data:
[tex](0.014)(6)(12.25*sin(37)) = (0.00949 kg*m^2)W[/tex]
solving for W, we get:
W = 65.25 rad/s
c)
The kinetic energy is calculated by the next equation:
K = [tex]\frac{1}{2}I_sW^2[/tex]
replacing the data:
k = [tex]\frac{1}{2}(0.00949)(65.25)^2[/tex]
k = 20.2021 J
