Problem 17: You have a horizontal grindstone (a disk) that is 87 kg, has a 0.39

Question

Problem 17: You have a horizontal grindstone (a disk) that is 87 kg, has a 0.39 m radius, is turning at 92 rpm (in the positive direction), and you press a steel axe against the edge with a force of 23 N in the radial direction Otheexpertta.com Part (a) Assuming that the kinetic coeflicient of friction between steel and stone is 0.20, calculate the angular acceleration of the grindstone in rad/s Numeric :A numeric value is expected and not an expression. Part (b) What is the number of turns, N, that the stone will make before coming to rest? Numeric : A numeric value is expected and not an expression

Explanation / Answer

here,

mass, m = 87 kg

radius , r = 0.39 m

a)

the angular accelration be alpha

I * alpha = r * F

0.5 * m * r^2 * alpha = r * ( uk * 23)

0.5 * 87 * 0.39 * alpha = 0.2 * 23

solving for alpha

alpha = 0.27 rad/s^2

b)

w = 92 rpm = 9.63 rad/s

let the angle covered be theta

2 * alpha * theta = w^2 - w0^2

2 * 0.27 * theta = 9.63^2

solving for theta

theta = 171.7 rad

the number of turns , N = theta /2pi = 27.3 turns

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