Question

A bar length of 3l and insignificant mass is suspended from a ceiling in horizontal position using two strings attached to ends of the bar. Two masses have mass m (left one) and 3m (right one) fixed to the bar. Their center are placed in one-third and two-thirds of the bar's length. The mass' sizes are very small compared to the bar length . then a lighting bolt strikes the right string severing its connection to the bar causing the system to spin clock wise. 1. Find and expression from the tension T in the left string while the bar rotates in terms of m and g. 2. what is the magnitude and direction of the change in angular momentum of the system after 0.25 seconds has eslapsed, if the m=0.125 kg and l=1m

Answer 1

The tension in the left string is (10m/3) × g, and the change in angular momentum is approximately 1.03125 N·m·s. The rotation is clockwise.

Expression for Tension T in the Left String:

The system is in rotational equilibrium, so the torque acting on the system is zero. The torque is given by the tension T in the left string and the weight of the masses.

The torque due to the mass m on the left side is (m/3) × g × l, and the torque due to the mass 3m on the right side is 3m × g × (2l/3). The total torque is:

T × l = (m/3) × g × l + 3m × g × (2l/3)

Solve for T:

T = (m + 9m)/3 × g

T = (10m/3) * g

So, the expression for tension T in the left string is (10m/3) × g.

Change in Angular Momentum:

The change in angular momentum (ΔL) is given by the torque (τ) applied over time (Δt):

ΔL = τ × Δt

The torque (τ) is provided by the tension in the left string and is equal to T × l.

ΔL = T × l × Δt

Substitute the expression for T:

ΔL = (10m/3) × g × l × Δt

Plug in the values:

ΔL = (10 × 0.125/3) × 9.8 × 1 × 0.25

ΔL ≈ 1.03125 N·m·s

The magnitude of the change in angular momentum is approximately 1.03125 N·m·s. Since the system is rotating clockwise, the direction of the change in angular momentum is clockwise.