Respuesta :
Answer:
The net torque is 0.0372 N m.
Explanation:
A rotational body with constant angular acceleration satisfies the kinematic equation:
[tex]\omega^{2}=\omega_{0}^{2}+2\alpha\Delta\theta [/tex] (1)
with ω the final angular velocity, ωo the initial angular velocity, α the constant angular acceleration and Δθ the angular displacement (the revolutions the sphere does). To find the angular acceleration we solve (1) for α:
[tex]\frac{\omega^{2}-\omega_{0}^{2}}{2\Delta\theta}=\alpha [/tex]
Because the sphere stops the final angular velocity is zero, it's important all quantities in the SI so 2.40 rev/s = 15.1 rad/s and 18.2 rev = 114.3 rad, then:
[tex]\alpha=-\frac{-(15.1)^{2}}{2(114.3)}=1.00\frac{rad}{s^{2}} [/tex]
The negative sign indicates the sphere is slowing down as we expected.
Now with the angular acceleration we can use Newton's second law:
[tex]\sum\overrightarrow{\tau}=I\overrightarrow{\alpha} [/tex] (2)
with ∑τ the net torque and I the moment of inertia of the sphere, for a sphere that rotates about an axle through its center its moment of inertia is:
[tex] I = \frac{2MR^{2}}{5}[/tex]
With M the mass of the sphere an R its radius, then:
[tex] I = \frac{2(1.85)(\frac{0.45}{2})^{2}}{5}=0.037 kg*m^2 [/tex]
Then (2) is:
[tex]\sum\overrightarrow{\tau}=0.037(-1.00)=0.037 Nm [/tex]
Answer:
D. 0.037 N m.
Explanation:
Given:
Diameter, d = 45 cm
Radius, r = d/2
= 0.225 m
Mass, m = 1.85 kg
Initial velocity, wi = 2.4 rev/s
θ = 18.2 rev
Final velocity, wf = 0 rev/s
Inertia of a uniform sphere, I = 2/5 × m × R^2
= 2/5 × 1.85 × 0.225^2
= 0.0375 kg.m^2
Using equation of angular motion,
wf^2 = wi^2 + 2ao × θ
Where,
ao = angular acceleration
ao = (2.4^2)/2 × 18.2
= 0.158 rev/s^2
Converting from rev/s^2 to rad/s^2,
0.158 rev/s^2 × 2pi rad/1 rev
= 0.993 rad/s^2
Torque, t = I × ao
= 0.0375 × 0.993
= 0.037 Nm
