Acircular saw blade with radius 0.175 m starts from rest and turns in a vertical plane with a constant angular acceleration of 2.00 rev/s^2. after the blade has turned through 155 rev, a small piece of the blade breaks loose from the top of the blade. after the piece breaks loose, it travels with a velocity that is initially horizontal and equal to the tangential velocity of the rim of the blade. the piece travels a vertical distance of 0.820 m to the floor.

(a) how far does the piece travel horizontally, from where it broke off the blade until it strikes the floor? express your answer with the appropriate units.

x = 11.23  m

Explanation:

For this interesting exercise, we must use angular kinematics, linear kinematics and the relationship between angular and linear quantities.

Let's reduce to SI system units

    θ = 155 rev (2pi rad / rev) = 310π rad

    α = 2.00rev / s2 (2pi rad / 1 rev) = 4π rad / s²

Let's look for the angular velocity at the time the piece is released, with starting from rest the initial angular velocity is zero (wo = 0)

    w² = w₀² + 2 α θ  

    w =√ 2 α θ

    w = √(2 4pi 310pi)

    w = 156.45  rad / s

The relationship between angular and linear velocity

    v = w r

    v = 156.45  0.175

    v = 27.38 m / s

In this part we have the linear speed and the height that it travels to reach the floor, so with the projectile launch equations we can find the time it takes to arrive

    y = t - ½ g t²

As it leaves the highest point its speed is horizontal

   y = 0 - ½ g t²

   t = √ (-2y / g)

   t = √ (-2 (-0.820) /9.8)

   t = 0.41 s

With this time we calculate the horizontal distance, because the constant horizontal speed

   x = vox t

   x = 27.38 0.41

   x = 11.23  m

d.7.5cm is your answer