Respuesta :
Answer:
w_f = m*V*cos(Q_n) / L*(m+M)
Explanation:
Given:
- mass of the putty ball m
- mass of the rod M
- Velocity of the ball V
- Length of the rod L
- Angle the ball makes before colliding with rod Q_n
Find:
What is the angular speed ωf of the system immediately after the collision,
Solution:
- We can either use conservation of angular momentum or conservation of Energy. We will use Conservation of angular momentum of a system:
L_before = L_after
- Initially the rod is at rest, and ball is moving with the velocity V at angle Q from normal to the rod. We know that the component normal to the rod causes angular momentum. Hence,
L_before = L_ball = m*L*V*cos(Q_n)
- After colliding the ball sicks to the rod and both move together with angular speed w_f
L_after = (m+M)*L*v_f
Where, v_f = L*w_f
L_after = (m+M)*L^2 * w_f
- Now equate the two expression as per conservation of angular momentum:
m*L*V*cos(Q_n) = (m+M)*L^2 * w_f
w_f = m*V*cos(Q_n) / L*(m+M)
Answer:
The angular speed [tex]\omega_f[/tex] of the system is [tex]\frac{m\times v \rm\times \cos\Theta}{(m+M)\times \rm L}=\omega_f[/tex]
Explanation:
- Mass of the putty ball m
-Mass of the rod M
-Velocity of the ball V
-Length of the rod L
-Angle the ball makes before colliding with rod [tex]\Theta[/tex].
By conservation of angular momentum,
[tex]\vec{L}=\vec{r}\times\vec{p}=\vec{r}\times m\vec{v}[/tex]
Initial angular momentum=Final angular momentum
[tex]L_i=L_f[/tex]
[tex]L_{initial} =m\times v\times \rm L\times \cos\Theta[/tex]
[tex]L_{final} =(m+M)\times \rm L^2\times \omega_f[/tex]
On substitution,we get
[tex]m\times v\times \rm L\times \cos\Theta=(m+M)\times \rm L^2\times \omega_f[/tex]
[tex]\frac{m\times v \rm\times \cos\Theta}{(m+M)\times \rm L}=\omega_f[/tex]
Hence,
The angular speed [tex]\omega_f[/tex] of the system is [tex]\frac{m\times v \rm\times \cos\Theta}{(m+M)\times \rm L}=\omega_f[/tex]
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