Answer:
2.37 Nm
Explanation:
The two balls are balanced at the center of gravity. By balancing the torque generated by the weight of each ball, we can locate the center of gravity. Next, we can solve for the net torque needed to stop the rotation of the rod-ball system using either angular impulse or kinematics and Newton's second law of motion.
First, we find the center of gravity by balancing the torques of the weight of the two balls. The torque is equal to the weight force times the distance from the center of gravity.
τ₁ = τ₂
m₁g₁ r₁ = m₂g₂ r₂
(4.0 kg) (9.8 m/s²) (x) = (2.0 kg) (9.8 m/s²) (3.0 m − x)
2x = 3.0 m − x
3x = 3.0 m
x = 1.0 m
The center of gravity is 1.0 m from the 4.0 kg ball.
Next, find the net torque required to stop the rotation of the system using one of the two methods mentioned.
The angular impulse on the system is equal to the net torque (τ) times the time (t). It is also equal to the moment of inertia (I) times the change in angular velocity (Δω). For a point mass, the moment of inertia is equal to the mass (m) times the square of the distance (r).
τ t = I Δω
τ t = (m₁ r₁² + m₂ r₂²) ω
τ = (m₁ r₁² + m₂ r₂²) ω / t
Convert rpm to rad/s.
ω = 17 rev/min × (2π rad/rev) × (1 min / 60 s)
ω = 1.78 rad/s
Plug in values:
τ = [(4.0 kg) (1.0 m)² + (2.0 kg) (2.0 m)²] (1.78 rad/s) / (9.0 s)
τ = 2.37 Nm
The angular acceleration of the system (α) is equal to the change in angular velocity (Δω) over time (t).
α = Δω / t
α = (1.78 rad/s) / (9.0 s)
α = 0.198 rad/s²
From Newton's second law of motion, the net torque (τ) on the system is equal to the moment of inertia (I) times the angular acceleration (α).
τ = Iα
τ = (m₁ r₁² + m₂ r₂²) α
τ = [(4.0 kg) (1.0 m)² + (2.0 kg) (2.0 m)²] (0.198 rad/s²)
τ = 2.37 Nm
