A circular saw blade with radius 0.110 m starts from rest and turns in a vertica

Question

A circular saw blade with radius 0.110 m starts from rest and turns in a vertical plane with a constant angular acceleration of 2.00 rev/s2. 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?

Explanation / Answer

here,

radius ,r = 0.11 m

accelration ,a = 2 rev/s^2

a = 12.56 rad/s^2

theta = 155 rev = 973.4 rad

let the final speed be w

w^2 - w0^2 = 2 * a * theta

w^2 - 0 = 2 * 12.56 * 973.4

w = 156.37 rad/s

the horizontal speed , v = r * w

v = 17.2 m/s

vertical distance , h = 0.82 m

let the time taken to reach the floor be t

h = u*t + 0.5 * g * t^2

0.82 = 0 + 0.5 * 9.8 * t^2

t = 0.41 s

the horizontal distance travelled , x = v * t

x = 7.04 m

the distance piece travel horizontally is 7.04 m

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