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8090 [49]
3 years ago
9

A uniform cylindrical grindstone has a mass of 10 kg and a radius of 12 cm. (a) What is the rotational kinetic energy of the gri

ndstone when it is rotating at 1.5×103rev/min? (b) After the grindstone’s motor is turned off, a knife blade is pressed against the outer edge of the grindstone with a perpendicular force of 5.0 N. The coefficient of kinetic friction between the grindstone and the blade is 0.80. Use the work energy theorem to determine how many turns the grindstone makes before it stops.
Physics
1 answer:
Drupady [299]3 years ago
6 0

Answer:

a) KE = 888.26J

b) N = 294.5 turns

Explanation:

For the kinetic energy:

KE = I/2*\omega_o^2

The inertia is:

I=m/2*R^2=0.072kg.m^2

So, the kinetic energy will be:

KE = 888.26J

Now, friction force is:

Ff = μ*N = 0.80*5N = 4N

The energy balance would be:

Kf - Ko = Wf    where Kf=0;   Ko = 888.26J;  and Wf is the work done by friction force.

Wf = -Ff*d = -Ff*N*2*π*R   where N is the amount of turns it gives.

Replacing these values into the energy balance:

0-888.26=-4*N*2*π*0.12

-888.26=-0.96*π*N

N=294.5 turns

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Explanation:

First to all, let's gather the data. We know that both rods, have the same length. Now, the expression to use here is the following:

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This is the equation that describes the relation between speed of a pulse and a force exerted on it.

the value of "u" is:

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Where m is the mass of the rod, and L the length.

Now, for the rod 1:

V1 = √F/u1 (1)

rod 2:

V2 = √F/u2 (2)

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Replacing (2) in (3):

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Now, let's solve the equation 4:

[1.4(√F/u2)]² = F/u1

1.96(F/u2) =F/u1

1.96F = F*u2/u1

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Now, replacing the expression of u into (5) we have the following:

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But we need m1/m2 so:

1.96m1 = m2

m1/m2 = 1/1.96

m1/m2 = 0.51

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3 years ago
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