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3 years ago

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A tradesman sharpens a knife by pushing it with a constant force against the rim of a grindstone. The 30-cm-diameter stone is sp

inning at 200 rpm and has a mass of 28 kg. The coefficient of kinetic friction between the knife and the stone is 0.2. The stone slows steadily to 180 rpm in 10 s of grinding.
a) What is the magnitude of the angular acceleration of the grindstone as it slows down?

b) With what force does the tradesman press the knife against the grindstone?

1 answer:

Lorico [155]3 years ago

7 0

Answer:

a. 0.21 rad/s2

b. 2.205 N

Explanation:

We convert from rpm to rad/s knowing that each revolution has 2π radians and each minute is 60 seconds

200 rpm = 200 * 2π / 60 = 21 rad/s

180 rpm = 180 * 2π / 60 = 18.85 rad/s

r = d/2 = 30cm / 2 = 15 cm = 0.15 m

a)So if the angular speed decreases steadily (at a constant rate) from 21 rad/s to 18.85 rad/s within 10s then the angular acceleration is

$\alpha = \frac{\Delta \omega}{\Delta t} = \frac{21 - 18.85}{10} = 0.21 rad/s^2$

b) Assume the grind stone is a solid disk, its moment of inertia is

$I = mR^2/2$

Where m = 28 kg is the disk mass and R = 0.15 m is the radius of the disk.

$I = 28*0.15^2/2 = 0.315 kgm^2$

So the friction torque is

$T_f = I\alpha = 0.315*0.21 = 0.06615 Nm$

The friction force is

$F_f = T_f/R = 0.06615 / 0.15 = 0.441 N$

Since the friction coefficient is 0.2, we can calculate the normal force that is used to press the knife against the stone

$N = F_f/\mu = 0.441/0.2 = 2.205 N$

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$tan \theta_1 = \dfrac{-[0.875\times 10.5 \times Sin 21.5^0]}{7.25\times 9.85 - (0.875\times 10.5 \times cos 21.5^0)}$

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$tan \theta_1 = 0.0536$

$\theta_1 =-3.066^0$

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$v = \dfrac{-m_2v_2sin\theta_2}{m_1 sin\theta_1}$

$v = \dfrac{-0.875 \times 10.5 sin21.5^0}{7.25 sin(-3.066^0)}$

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solution

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