A circular saw blade with radius 0.135 m starts from rest and turns in a vertica

Question

A circular saw blade with radius 0.135 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.

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

Explanation / Answer

155 rev = 973.894 rad

2 rev/s^2 = 12.5664 rad/sec^2

v^2 = u^2 + 2as

v^2 = 0 + 2 * 12.5664 * 973.894

v = 156.45 rad/sec

linear speed = angular speed * radius

linear speed = 156.45 * 0.135

linear speed = 21.12075 m/s

time it'll take to fall 0.82 m

0.82 = 0.5 * 9.8 * t^2

t = 0.409 sec

horizontal distance covered in 0.409 sec = 0.409 * 21.12075

horizontal distance covered = 8.6383 m

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