A circular saw blade with radius 0.185m starts from rest and turns in a vertical

Question

A circular saw blade with radius 0.185m 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.

How far does the piece travel horizontally, from where it broke off the blade until it strikes the floor?

Explanation / Answer

here,

radius , r = 0.185 m

angular accelration ,a = 2 rev/s^2

a = 12.28 rad/s^2

theta = 155 rev = 973.4 rad

let the final angular speed be w

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

w^2 - 0 = 2 * 12.28 * 973.4

w = 154.62 rad/s

tangential speed , v = r * w

v=0.185*154.62

v = 28.60 m/s

let the time taken to reach ground be t

h0 = 0.82 = u*t + 0.5 * g *t^2

0.82 = 0 + 0.5 * 9.8 * t^2

t = 0.41 s

the horizontal distance , x = v * t = 28.60 * 0.41

x = 11.73 m

the horizontal distance travelled by the broke off blade is 11.73 m

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