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Answer
given,
L(t) = 10 - 3.5 t
mass of particle = 2 Kg
radius of the circle = 3.1 m
a) torque
τ = 
τ = 
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 = 
ω = 
ω = 
A = 0.52 rad/s B = -0.182 rad/s²
Answer:
Acceleration = 311.2 Km/hr²
Explanation:
Given: Radius of the Orbit r= 3.56 × 10⁶ km
Period of the orbit = 28 days = 672 hrs
Sol: We have Fc = MV²/r
⇒M ac = MV²/r
⇒ac = V²/r
First we have to Speed V so for this we have to find the circumference ( distance covered by the moon in one orbit)
⇒ Circumference= 2 π r
= 2 × 3.13149 × 3.56 × 10⁶ km
= 22,368,139.69 Km
Now Speed = Distance /time
Speed = 22,368,139.69 Km / 672 hrs
Speed V = 33,285.92 Km/Hr
Now
ac = V²/r = (33,285.92 Km/hrs)² / 3.56 × 10⁶ km
ac = 311.2 Km/hr²