Respuesta :
Answer:
b. 29.2 rev/min
Explanation:
- Assuming no external torques acting during the process, total angular momentum must be conserved, as follows:
[tex]L_{0} = L_{f} (1)[/tex]
- The initial angular momentum L₀, can be expressed as follows:
[tex]L_{0} = I_{0} * \omega_{0} (2)[/tex]
where I₀ = initial moment of inertia = moment of inertia of the disk +
moment of inertia of the cylinder and ω₀ = initial angular velocity =
30.0 rev/min.
- Replacing by the values, we get:[tex]I_{0} = \frac{1}{2} * m_{d} *r_{d} ^{2} + \frac{1}{2}* m_{c} *r_{c} ^{2} = 0.2 kg*m2 +9e-4 kg*m2 = 0.2009 kg*m2 (3)[/tex]⇒ L₀ = I₀* ω₀ = 0.2009 kg*m² * 30.0 rev/min = 6.027 kg*m²*rev/min
- The final angular momentum can be written as follows:
[tex]L_{f} = I_{f} * \omega_{f} (4)[/tex]
where If = final moment of inertia = moment of the inertia of the solid
disk + moment of inertia of the clay flattened on a disk, and ωf = final
angular velocity.
- Replacing by the values, we get:
[tex]I_{f} = \frac{1}{2} * m_{d} *r_{d} ^{2} + \frac{1}{2}* m_{fd} *r_{fd} ^{2} = 0.2 kg*m2 +6.4e-3 kg*m2 = 0.2064 kg*m2 (5)[/tex]
⇒ Lo =Lf = If*ωf
- Replacing (2) in (1), and solving for ωf, we get:
[tex]\omega_{f} = \frac{L_{o}}{I_{f} } = \frac{6.027kg*m2*rev/min}{0.2064kg*m2} = 29.2 rev/min (6)[/tex]
The rate of change of angular displacement is defined as angular speed. The final angular speed of the wheel will be 29.2 rev/min.
What is angular speed?
The rate of change of angular displacement is defined as angular speed. is stated as follows:
ω = θ t
Where,
θ is the angle of rotation,
t is the time
ω is the angular speed
The given data in the problem is
m is the mass of wheel = 10.0 kg
r₁ is the radius of disk = 20.0 cm=0.2
M is the mass of clay= 2.0 kg
R is the radius of cylinder = 3.0cm
[tex]\rm \omega_i[/tex] is the initial rotational speed =30.0 rev/min
r₂ is the final radius of disk= 8.0 cm.
[tex]\rm \omega_f[/tex] is the initial rotational speed=?
When the external torques act on the body is zero the total angular momentum must be conserved, as follows:
Initial momentum = Final momentum
[tex]\rm L_0=L_f[/tex]
The value of the initial angular momentum L₀ is found by
I₀ = initial moment of inertia = moment of inertia of the disk +moment of inertia of the cylinder
[tex]\rm I_0= \frac{1}{2}m_dr_d^2+ \frac{1}{2}m_cr_c^2\\\\ \rm I_0= \frac{1}{2}\times 10\times (0.2)^2+ \frac{1}{2}\times m_2(0.03)^2[/tex]
[tex]\rm L_0 = I_0\times \omega_0\\\\ L_0 = 2009\times 30\\\\ \rm L_0 =6.027 \;kgm^2rev/min[/tex]
The value of the final angular momentum [tex]I_f[/tex] is found by
[tex]\rm I_f= \frac{1}{2}m_dr_d^2+ \frac{1}{2}m_fr_fd^2\\\\ \rm I_0= 0.2064 m_2[/tex]
[tex]\rm I_f[/tex] is the final moment of inertia = moment of the inertia of the solid disk + moment of inertia of the clay flattened on a disk.
[tex]L_0 =L_f = I_f \times \omega_f[/tex]
[tex]\rm \omega_f=\frac{L_0}{I_f} \\\\ \rm \omega_f=\frac{6.027 m_2}{0.2064 m_2} \\\\ \rm \omega_f= 29.2\; rev/min[/tex]
Hence the final angular speed of the wheel will be 29.2 rev/min.
To learn more about the angular speed refer to the link;
https://brainly.com/question/9684
