You have a grindstone (a disk) that is 76.0 kg, has a 0.430-m radius, and is tur

Question

You have a grindstone (a disk) that is 76.0 kg, has a 0.430-m radius, and is turning at 83.0 rpm, and you press a steel axe against it with a radial force of 10.0 N.

(a) Assuming the kinetic coefficient of friction between steel and stone is 0.50, calculate the angular acceleration (in rad/s2) of the grindstone. (Indicate the direction with the sign of your answer.) -0.31 Correct: Your answer is correct. rad/s2

I was able to work out the first part, but I am struggling with the second.

(b) How many turns (in rev) will the stone make before coming to rest?

rev?

Explanation / Answer

a)

radial force is Fnet = m*a = 10

accelaration is a = 10/m = 10/76 = 0.131 m/s^2

but a = r*alpha

angular accelaration is alpha = a/r = 0.131/0.43 = 0.31 rad/s^2

alpha = -0.31 rad/s^2

b) initial angular velocity is wo = 83 rpm = 83*(2*3.142/60) = 8.69 rad/s

final angular velocity is w = 0 rad/s

alpha =0.31 rad/s^2

then

using

w^2-wi^2 = 2*alpha*theta

0^2-8.69^2 = -2*0.31*theta

theta = 121.8 rad

no.of turns are 121.8/(2*pi*0.43) = 45 rev

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