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A 13.0 kg bucket of water of is suspended by a rope wrapped around a pulley, which is just a solid cylinder 0.300 m in diameter with mass 17.0 kg. The cylinder is pivoted on a frictionless axle through its center. The bucket is released from rest at the top of a well and falls 12.0 m to the water. Neglect the weight of the rope.

(a) What is the tension in the rope while the bucket is falling?

50.367 N

(b) With what speed does the bucket strike the water?

11.925 m/s

(c) What is the time of fall?

2.013 s

(d) While the bucket is falling, what is the force exerted on the cylinder by the axle?

Attempt

F=Mg-Ma

then implied:

a=(F-Mg)/M

which seems unsound, it should have said:

a = (Mg - F)/M

----------------------

Now, let's begin and use the previous gentleman's notation:

Mass of bucket=M

mass of cylinder=m

Tension force=F

So, regarding the bucket:

F + Ma = Mg

F = Mg - Ma

a = (Mg - F)/M

Also, regarding the cylinder:

T = Iα

∴

(By the dot product of torque, radial vector crossed with force, r cross F, which is Frsinθ for θ = 90 degrees)

T = Iα = Fr = (1/2)(mr2)(a/r)

where α = a/r

And T = Fr because that's what torque is, just the radial distance times the force.

So

a = 2F/m

and we also have:

a = (Mg-F)/M

So 2F/m = (Mg-F)/M

2F = mg - mF/M

2F+(mF)/M = mg

F(2+m/M) = mg

F = mg/(2+m/M)

F = (167 kg*m/s2)/(2+1.31kg/kg)

F = 50.367 N

That is to say, the tension that the rope induces on the bucket is 50.5 Newtons in the upward direction (positive).

--------------

So we have solved Part (A)

F = 50.367 N

---------------

We know that:

F = Mg-Ma

So

a = (Mg - F)/M

where F = 50.367 N

So

a = [(13.0kg)(9.8m/s2) - 50.367N]/13.0kg

a = (127N - 50.367N)/13.0kg

a = 5.925 m/s2

--------------------

Now that we have a, let's find t,

a = a

v = at + vyo

y = yo + vyot + (1/2)ayt2

Counting positive y to be downward, (our acceleration will be positive then)

12.0 m = 0 + 0t + (1/2)(5.925m/s2)t2

t2 = (12.0m)/(2.963 m/s2)

t2 = 4.051 s2

t = 2.013 s

That is to say, it takes the bucket 2.10 seconds to reach the water surface below.

--------------

So we have solved Part (C)

t = 2.013 s

---------------

a = a

v = at + C

But C, the constant of integration, is the initial velocity, is zero.

so

v = at

where t =2.013 s

a = 5.925 m/s2

so

v = (5.925m/s2)(2.013s)

v = 11.925 m/s

That is to say, the bucket is travelling at 11.925 m/s when it reaches the surface of the water.

--------------

So we have solved Part (B)

v = 11.925 m/s

---------------

The force exerted on the cylinder by the axle is just the torque (which is in the z direction, like a pole coming out of the page).

We know that the torque is:

T = Iα

where moment of Inertia for a cylinder is

I = (1/2)mr2

and the angular acceleration is

α = a/r

with a = 5.925m/s2

r = 0.150 m

So

T = Iα

T = (1/2)(17.0kg)(0.150m)2[5.925m/s2]/(0.150m)

T = (8.5kg)(0.150m)(5.925m/s2)

T = 7.555 N*m

or

T = 7.555 J

T= Fr

F=T/r

F= 7.555/0.150

F= 50.367 N

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# Homework Help: Bucket of water of is suspended by a rope wrapped around a pulley

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