A safety table saw uses a braking system to stop the saw blade from rotating if

Question

A safety table saw uses a braking system to stop the saw blade from rotating if it senses that the blade is touching the user's finger. To avoid injuring the finger, the blade must be stopped within the time it would take the blade to rotate about 1/20 of a turn at full speed. Assume the saw blade is essentially a flat disk with mass M and radius R, and rotates once in time T at full speed. (a) What is the magnitude |F| of the constant force that the brake must exert on the rim when tripped in terms of these quantities? (b) If the saw blade has a radius of 13 cm, a mass of 600 g, and rotates at 3000 rpm, what is |F|? (The braking system used in safety saws exerts this enormous force by jamming a soft aluminum block into the blade's teeth. The force is exerted as a result of the teeth burying themselves in the block.)

Explanation / Answer

(A) Stopping time = T /20 sec

initial angular velocity. w0 = 2pi / T

Applying wf = w0 + alpha t

0 = (2 pi / T) + alpha (T / 20)

alpha = - 40 pi / T^2

I = M R^2 /2

and torque = I alpha

R F = (M R^2 / 2) (40 pi / T^2)

F = 20 pi M R / T^2

(B) T = 2pi / w

w = 3000 rpm

w = 3000 x 2pi rad / 60 s = 314 rad/s

T = 0.02 sec   

F = (20 x pi x 0.600 x 0.13) / 0.02^2

F = 12252 N

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