Explanation:
The given data is as follows.
= 57 kg,
= 79 kg
= 6.5 m,
= (6.5 - 1.9) m = 4.6 m
(a) The sum of torque ends about far end is as follows.
= 0
= 0
T = 828 N
Therefore, 828 N is the tension in the cable closer to the painter.
(b) Now, we will calculate the sum about close ends as follows.
= 0
T= 506 N
Therefore, 506 N is the tension in the cable further from the painter.
Complete question:
A solid disk rotates in the horizontal plane at an angular velocity of 0.067 rad/s with respect to an axis perpendicular to the disk at its center. The moment of inertia of the disk is 0.18 kg.m2. From above, sand is dropped straight down onto this rotating disk, so that a thin uniform ring of sand is formed at a distance of 0.40 m from the axis. The sand in the ring has a mass of 0.50 kg. After all the sand is in place, what is the angular velocity of the disk?
Answer:
The angular velocity of the disk is 0.0464 rad/s
Explanation:
Given;
initial angular velocity of disk, ωi = 0.067 rad/s
initial moment of inertia of the disk, I₁ = 0.18 kg.m²
radius of sand on the disk, R = 0.40 m
mass of sand, m = 0.50 kg
Initial angular momentum = Final angular momentum
Moment of inertia of sand ring = MR²
Therefore, the angular velocity of the disk is 0.0464 rad/s