Respuesta :
Answer:
w_f = 1.0345 rad/s
Explanation:
Given:
- The mass of the solid cylinder M = 45 kg
- Radius of the cylinder R = 0.44 m
- The mass of the particle m = 3.6 kg
- The initial speed of cylinder w_i = 0 rad/s
- The initial speed of particle V_pi = 3.3 m/s
- Mass moment of inertia of cylinder I_c = 0.5*M*R^2
- Mass moment of inertia of a particle around an axis I_p = mR^2
Find:
- What is the magnitude of its angular velocity after the collision?
Solution:
- Consider the mass and the cylinder as a system. We will apply the conservation of angular momentum on the system.
L_i = L_f
- Initially, the particle is at edge at a distance R from center of cylinder axis with a velocity V_pi = 3.3 m/s contributing to the initial angular momentum of the system by:
L_(p,i) = m*V_pi*R
L_(p,i) = 3.6*3.3*0.44
L_(p,i) = 5.2272 kgm^2 /s
- While the cylinder was initially stationary w_i = 0:
L_(c,i) = I*w_i
L_(c,i) = 0.5*M*R^2*0
L_(c,i) = 0 kgm^2 /s
The initial momentum of the system is L_i:
L_i = L_(p,i) + L_(c,i)
L_i = 5.2272 + 0
L_i = 5.2272 kg-m^2/s
- After, the particle attaches itself to the cylinder, the mass and its distribution around the axis has been disturbed - requires an equivalent Inertia for the entire one body I_equivalent. The final angular momentum of the particle is as follows:
L_(p,f) = I_p*w_f
- Similarly, for the cylinder:
L_(c,f) = I_c*w_f
- Note, the final angular velocity w_f are same for both particle and cylinder. Every particle on a singular incompressible (rigid) body rotates at the same angular velocity around a fixed axis.
L_f = L_(p,f) + L_(c,f)
L_f = I_p*w_f + I_c*w_f
L_f = w_f*(I_p + I_c)
-Where, I_p + I_c is the new inertia for the entire body = I_equivalent that we discussed above. This could have been determined by the superposition principle as long as the axis of rotations are same for individual bodies or parallel axis theorem would have been applied for dissimilar axes.
L_i = L_f
5.2272 = w_f*(I_p + I_c)
w_f = 5.2272/ R^2*(m + 0.5M)
Plug in values:
w_f = 5.2272/ 0.44^2*(3.6 + 0.5*45)
w_f = 5.2272/ 5.05296
w_f = 1.0345 rad/s
The magnitude of the angular velocity of the solid cylinder wuth given parameters after collision is; ω_f = 1.0345 rad/s
What is the magnitude of the angular velocity?
We are given:
Mass of the solid cylinder; M = 45 kg
Radius of the cylinder; R = 0.44 m
mass of the particle; m = 3.6 kg
initial angular speed of cylinder; ω_i = 0 rad/s
initial speed of particle V_pi = 3.3 m/s
Mass moment of inertia of cylinder I_c = 0.5*M*R²
Mass moment of inertia of a particle around an axis; I_p = mR²
ApplyIing the conservation of angular momentum on the system, we have; L_i = L_f
Thus;
L_(p,i) = m × V_pi × R
L_(p,i) = 3.6 × 3.3 × 0.44
L_(p,i) = 5.2272 kg.m²/s
The cylinder is stationary and as such; ω_i = 0:
Thus; L_(c,i) = 0 kg.m²/s
The initial angular momentum of the system is;
L_i = L_(p,i) + L_(c,i)
L_i = 5.2272 + 0
L_i = 5.2272 kg.m²/s
From the parallel axis theorem, we have;
L_i = L_f
5.2272 = ω_f × (I_p + I_c)
ω_f = 5.2272/(R²(m + 0.5M))
Thus;
ω_f = 5.2272/0.44² × (3.6 + 0.5*45)
ω_f = 5.2272/5.05296
ω_f = 1.0345 rad/s
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