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

1. Jan 3, 2010

### squintyeyes

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 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

2. Jan 3, 2010

### Staff: Mentor

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. Jan 3, 2010

### squintyeyes

so is it just the force of tension. that means the answer to a and d are the same.

4. Jan 3, 2010

### Staff: Mentor

No. You're overlooking one of the forces acting on the cylinder.

5. Jan 3, 2010

### squintyeyes

oh would i do
force of gravity + tension = force on cylinder
mg + 50.367 =
17(9.8) + 50.367=
216.967 N

6. Jan 3, 2010

Good.

7. Jan 3, 2010

### squintyeyes

thank you for your help