Respuesta :
Answer
given,
L(t) = 10 - 3.5 t
mass of particle = 2 Kg
radius of the circle = 3.1 m
a) torque
τ = [tex]\dfrac{dL}{dt}[/tex]
τ = [tex]\dfrac{d}{dt}(10 - 3.5 t)[/tex]
τ = -3.5 N.m
Particle rotates clockwise as i look down the plane. Hence, its angular velocity is downward.
L decreases the angular acceleration upward. so, net torque is upward.
b) Moment of inertia of the particle
I = m R^2
I = 2 x 3.1²
I = 19.22 kg.m²
L = I ω
ω = [tex]\dfrac{L}{I}[/tex]
ω = [tex]\dfrac{10 - 3.5 t}{19.22}[/tex]
ω = [tex]0.520 - 0.182 t[/tex]
A = 0.52 rad/s B = -0.182 rad/s²
Answer:
Explanation:
mass, m = 2 kg
radius, r = 3.1 m
L(t) = 10 - 3.5 t
(a) [tex]\tau =\frac{dL}{dt}[/tex]
So, differentiate angular momentum with respect to time
τ = - 3.5 Nm
As the particle is rotating clockwise, so the angular velocity is downward and the angular momentum decreases so torque is upward.
(b) moment of inertia, I = m R²
I = 2 x 3.1 x 3.1 = 19.22 kgm^2
L = I x ω
where, ω is the angular velocity
ω = (10 - 3.5 t) / 19.22
ω = 0.520 - 0.182 t
So, by comparison A = 0.520 rad/s
B = - 0.182 rad/s²
