Answer:
Torque = –207.4 Nm
Explanation:
Given M = 3.2kg, r = 5.4m, α = –12rad/s² (it is slowing down)
Torque = I × α
α = angular acceleration
I = moment of inertia
I = MR² for a circular hoop
Torque = 3.2×5.4×(– 12)
Torque = –207.4 Nm
Explanation:
(a) Draw a free body diagram of the cylinder at the top of the loop. At the minimum speed, the normal force is 0, so the only force is weight pulling down.
Sum of forces in the centripetal direction:
∑F = ma
mg = mv²/RL
v = √(g RL)
(b) Energy is conserved.
EE = KE + RE + PE
½ kd² = ½ mv² + ½ Iω² + mgh
kd² = mv² + Iω² + 2mgh
kd² = mv² + (m RC²) ω² + 2mg (2 RL)
kd² = mv² + m RC²ω² + 4mg RL
kd² = mv² + mv² + 4mg RL
kd² = 2mv² + 4mg RL
kd² = 2m (v² + 2g RL)
d² = 2m (v² + 2g RL) / k
d = √[2m (v² + 2g RL) / k]
Mass of the bird(m) = 150 g = 0.15 kg
Speed (v) = 10 m/s
Kinetic Energy = 
= 7.5 J
Altitude (h) = 15 m
Gravitational Potential Energy = (0.15)(9.81)(15) = 22.0725 J
Mechanical Energy = Kinetic Energy + Potential Energy = 7.5 + 22.0725
= 29.5725 J
