Question

Two metal disks are welded together and are mounted on a frictionless axis through their common centers. One disk has a radius R1 = 2.60 cm and mass M1 = 0.810 kg and the other disk has a radius R2 = 5.00 cm and mass M2 = 1.58 kg . (Figure 1) A string of negligible mass is wrapped around the smaller disk, and a 1.50-kg block, is suspended from the free end of the string. The block is released from rest a distance of 2.10 m above the ground. What is the block's speed just before it hits the ground?

Answer 1

The Inertia is 22. 488 kg. m² and the speed just before it hits the ground is 6. 4 m/s

How to determine the inertia

Using the formula:

I = 1/2 M₁R₁² + 1/2 M₂R₂²

Where I = Inertia

I = 1/2 * 0.810* (2. 60)² + 1/2 * 1. 58 * (5)²

I = 1/2 * 5. 476 + 1/2 * 39. 5

I = 2. 738 + 19. 75

I = 22. 488 kg. m²

To determine the block's speed, use the formula

v = $\sqrt{2gh}$

v = $\sqrt{2* 10 * 2. 10}$

v = $\sqrt{42}$

v = 6. 4 m/s

Therefore, the Inertia is 22. 488 kg. m² and the speed just before it hits the ground is 6. 4 m/s

Learn more about law of inertia here:

brainly.com/question/10454047

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