Bucket of water of is suspended by a rope wrapped around a pulley

In summary, the tension in the rope while the bucket is falling is 50.367 Newtons. The bucket strikes the water at 11.925 m/s, and the force exerted on the cylinder by the axle is 7.555 Newtons.
  • #1
squintyeyes
45
0
how do you do the last part.

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 traveling 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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  • #2
squintyeyes said:
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).
Since the axle is frictionless, it exerts no torque. (If it did exert a torque, your previous work would be incorrect since it ignored it.)

Hint: Consider the forces acting on the cylinder. What must be the net force?
 
  • #3
so is it just the force of tension. that means the answer to a and d are the same.
 
  • #4
squintyeyes said:
so is it just the force of tension. that means the answer to a and d are the same.
No. You're overlooking one of the forces acting on the cylinder.
 
  • #5
oh would i do
force of gravity + tension = force on cylinder
mg + 50.367 =
17(9.8) + 50.367=
216.967 N
 
  • #6
Good.
 
  • #7
thank you for your help
 

1. How does a bucket of water suspended by a rope wrapped around a pulley stay in place?

The bucket of water stays in place because the pulley distributes the weight of the bucket and water evenly, making it easier to lift and keep suspended in the air.

2. What is the purpose of using a pulley for a bucket of water?

The pulley allows for the weight of the bucket and water to be distributed, making it easier to lift. It also allows for the bucket to be moved horizontally or vertically without having to carry the weight.

3. How does the direction of the pulley affect the movement of the bucket of water?

The direction of the pulley affects the movement of the bucket of water by changing the direction of the force needed to lift the bucket. For example, a fixed pulley changes the direction of the force needed to lift the bucket from downward to upward.

4. Can a pulley system be used to lift heavier objects than a human can normally lift?

Yes, a pulley system can be used to lift heavier objects than a human can normally lift because it distributes the weight and reduces the amount of force needed to lift the object.

5. How does friction affect the movement of a bucket of water suspended by a rope wrapped around a pulley?

Friction can affect the movement of a bucket of water suspended by a rope wrapped around a pulley by creating resistance and making it more difficult to lift the bucket. Lubricating the pulley and rope can help reduce friction and make the movement smoother.

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