Answer:
D) 0.0372 N m
Explanation:
r = 45/2 cm = 22.5 cm = 0.225 m
As 1 revolution = 2π rad we can convert to radian unit
2.4 rev/s = 2.4 * 2π = 15.1 rad/s
18.2 rev = 18.2 * 2π = 114.35 rad
We can calculate the angular (de)acceleration using the following equation of motion
[tex]-\omega^2 = 2\alpha \theta [/tex]
[tex]- 15.1^2 = 2*\alpha * 114.35[/tex]
[tex]\alpha = \frac{-15.1^2}{2*114.35} = -0.994 rad/s^2[/tex]
The moment of inertia of the solid uniform sphere is
[tex]2mr^2/5 = 2*1.85*0.225^2/5 = 0.0375 kgm^2[/tex]
The net torque acting on this according to Newton's 2nd law is
[tex]T = I\alpha = 0.0375 * 0.994 = 0.0372 Nm[/tex]
Answer:
(D) The net torque acting on this sphere as it is slowing down is closest to 0.0372 N.m
Explanation:
Given;
mass of the solid sphere, m = 1.85 kg
radius of the sphere, r = ¹/₂ of diameter = 22.5 cm
initial angular velocity, ω = 2.40 rev/s = 15.08 rad/s
angular revolution, θ = 18.2 rev = 114.37 rad
Torque on the sphere, τ = Iα
Where;
I is moment of inertia
α is angular acceleration
Angular acceleration is calculated as;
[tex]\omega_f^2 = \omega_i^2 +2 \alpha \theta\\\\0 = 15.08^2 + (2*114.37)\alpha\\\\\alpha = \frac{-15.08^2}{(2*114.37)} = -0.994 \ rad/s^2\\\\\alpha = 0.994 \ rad/s^2 \ (in \ opposite \ direction)[/tex]
moment of inertia of solid sphere, I = ²/₅mr²
= ²/₅(1.85)(0.225)²
= 0.03746 kg.m²
Finally, the net torque on the sphere is calculated as;
τ = Iα
τ = 0.03746 x 0.994
τ = 0.0372 N.m
Therefore, the net torque acting on this sphere as it is slowing down is closest to 0.0372 N.m
