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 2.50 rev/s2 . After the blade has turned through 145 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 0.700 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

angular acceleraiton , a = 2.5 rev/s^2

a = 15.71 rad/s^2

angular distance ,theta = 145 revs = 910.9 rads

radius of person , r = 0.110 m

final angular velocity , w = sqrt(2 * a * theta)

w = sqrt(15.71 * 910.9 * 2)

w = 169.2 rad/s

horizontal velocity = r * w

horizontal velocity = 0.11 * 169.2

horizontal velocity = 18.61 m/s

Now , time of flight , t = sqrt(2*h/g)

t = sqrt(2 * .7/9.8)

t = 0.38 s

Now , horizontal distance , d = 0.38 * 18.61

horizontal distance , d = 7.04 m

the horizontal distance travelled by the tip is 7.04 m

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