A circular saw blade with radius 0.140 m starts from rest and turns in a vertica

Question

A circular saw blade with radius 0.140 m starts from rest and turns in a vertical plane with a constant angular acceleration of 3.25 rev/s^2 . After the blade has turned through 160 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 0.775 m to the floor.

How far does the piece travel horizontally, from where it broke off the blade until it strikes the floor?
("x"-"x_o") Value & Units.

Explanation / Answer

Using = 0 t + 1/2t2

=>160*2 = 0 + 1/2 *3.25*2 t2

=>t2 = 98.461 s2

=> t= 9.922 s

Using =0 + t => = 0 + 3.25 * 9.922 = 32.25 rad/s

As v= r => v= 0.140* 32.25 = 4.515 m/s

Using equations of projectile motion,

s(vertical)=ut+1/2 at12 => 0.775= 0 + 1/2*9.8*t12 (here a=g)

t1 = 0.397 s

s(horizontal)=vt= 4.515*0.397 = 1.795 m

here s= x-x0

NOTE: Please mark me LIFESAVER...

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