Respuesta :
Answer:
Explanation:
Given that,
Height h = 29mm = 0.029m
Makes an angle of θ = 19.0°
Then, the radius is
h = rCosθ
r = h/Cosθ
r = 0.029/Cos19
r = 0.031m
Linear speed v = 1.45m/s
The relationship between linear speed and angular speed is
v = ωr
ω = v/r
ω = 1.45/0.031
ω = 47.28 rad/sec
Angular momentum is given as
L = Iω
Therefore, I = L / ω
So, the question is not complete check attachment for the first part of the question, from the first part we will get the angular momentum
Given that,
r = 3.46mm = 0.00346m
Tension in string F = 3.15N
Time taken t = 0.62s.
Then, Torque is given as
T = dL/dt
Tdt = dL
Integrate both sides
T•t + C = L(t)
At t = 0 L=0
Therefore C = 0
T•t = L(t)
The torque from the two string is given as
T = 2Fr
Therefore,
L(t) = 2•F•r•t
L(t) = 2 × 3.15 × 0.00346 × 0.62
L(0.62) = 0.0135 kgm²/s
Then, back to our question,
I = L / ω
I = 0.0135/47.28
I = 2.86 × 10^-4 kgm²
The application of the force on the axle causes the top to rotate with a
moment of inertia that can be found from the torque's magnitude.
- The top's moment of inertia is approximately 9.307044 × 10⁻⁵ kg·m².
Reasons:
The given parameter are;
Height of the top above the ground, h = 29.0 mm = 0.029 m
Angle made by the outer top with the axis of rotation = 19.0°
The tangential speed at point P = 1.45 m/s
Radius of the axle = 3.46 mm
Applied force on two strings to turn axle = 3.15 N each
Time of application of the force, t = 0.62 s
Required:
The moment of inertia of the top
Solution:
[tex]tan (\theta) = \dfrac{Radius}{Height}[/tex]
Radius of point P = 0.029 × tan(19°) ≈ 0.0099855008
[tex]Angular \ velocity, \ \omega = \dfrac{1.45}{0.0099855008} \approx 145.21[/tex]
Angular momentum, L = I·ω
[tex]Torque, \ T = \dfrac{L}{t}[/tex]
Where
L = T·t
T = Force × Radius
Applied force, F = 2 × 3.15 N (equal forces applied by pulling two ropes in
opposite direction).
Radius = The radius of the axle = 0.00346 m
Torque, T = 2 × 3.15 N × 0.00346 m = 0.021798 N·m
L = T × t
Therefore;
The angular momentum, L = 0.021798 N·m × 0.62 s = 0.01351476 N·m·s
0.01351476 N·m·s = 0.01351476 kg·m²/s
[tex]Moment \ of \ inertia, \ I = \mathbf{\dfrac{L}{\omega}}[/tex]
Therefore;
[tex]Top's \ moment \ of \ inertia, \ I = \dfrac{0.01351476 \ kg \cdot m^2/s}{145.21 \ rad/s} \approx 9.307044 \times 10^{-5} \ kg\cdot m^2[/tex]
- The top's moment of inertia, I ≈ 9.307044 × 10⁻⁵ kg·m²
Learn more here:
https://brainly.com/question/14460640
