Rotational ACC and Gravitational ACC

[DameLight]

solved thank you : )
1. Homework Statement
A bucket of water of mass 14.2 kg is suspended by a rope wrapped around a windlass, that is a solid cylinder with diameter 0.350 m with mass 12.8 kg . The cylinder pivots on a frictionless axle through its center. The bucket is released from rest at the top of a well and falls a distance 10.7 m to the water. You can ignore the weight of the rope.

no friction
m =14.2 kg
r = 1/2 0.350 m
M = 12.8 kg
Δy = 10.7 m
g = 9.8 m/s²

Homework Equations

t = Iα
F = ma
I = 1/2 mr²
a = rα

The Attempt at a Solution

What is the tension in the rope while the bucket is falling?[/B]

F = ma
the sum of the forces in the y direction for the bucket is (T - mg) so
T - mg = ma
T = ma + mg

a is unknown

t = F x r = Iα
since the string is pulling on the windlass and its mass is negligible (F = T) so
t = T x r = Iα
I = 1/2 Mr²
substitute this into the equation and simplify (did this for a previous question without values to find α)
T = 1/2 Mrα

set equations 1 and 2 equal to each other and solve for a
ma + mg = 1/2 Mrα
ma = 1/2 Mrα - mg

α is unknown

a = -rα
α = -a/r
substitute into equation 3
ma = 1/2 Mr(-a/r) - mg
ma = -1/2 Ma - mg
ma + 1/2 Ma = - mg
a(m + 1/2 M) = - mg
a = - (mg)/(m + 1/2 M)
a = - (14.2 kg*9.8 m/s²)/(14.2 kg + 1/2*12.8 kg)
a = - 6.76 m/s²

substitute back into equation 1
T = ma + mg
T = 14.2 kg* -6.76 m/s² + 14.2 kg*9.8 m/s²
T = 43.17 N

This is the correct answer : ) thank you for your help

conclusion:
reason for mistake,
1. sign mistake in ( T = ma - mg ) fixed to ( T = ma + mg )
2. a = -rα ; the sign conventions for this equation were wrong. I didn't really understand this equation, so I will study it more so that I don't make this mistake again.

 

[SammyS]

Homework Statement

A bucket of water of mass 14.2 kg is suspended by a rope wrapped around a windlass, that is a solid cylinder with diameter 0.350 m with mass 12.8 kg . The cylinder pivots on a frictionless axle through its center. The bucket is released from rest at the top of a well and falls a distance 10.7 m to the water. You can ignore the weight of the rope.

no friction
m =14.2 kg
r = 1/2 0.350 m
M = 12.8 kg
Δy = 10.7 m
g = 9.8 m/s²

Homework Equations

t = Iα
F = ma
I = 1/2 mr²
a = rα

The Attempt at a Solution

What is the tension in the rope while the bucket is falling?[/B]

F = ma
the sum of the forces in the y direction for the bucket is (T - mg) so
T - mg = ma
T = ma - mg $\ \ \$ There is a sign error here.

a is unknown

t = F x r = Iα
since the string is pulling on the windlass and its mass is negligible (F = T) so
t = T x r = Iα
I = 1/2 Mr²
substitute this into the equation and simplify (did this for a previous question without values to find α)
T = 1/2 Mrα

set equations 1 and 2 equal to each other and solve for a
ma - mg = 1/2 Mrα
ma = 1/2 Mrα + mg

α is unknown

a = rα
α = a/r
substitute into equation 3
ma = 1/2 Mr(a/r) + mg
ma = 1/2 Ma + mg
ma - 1/2 Ma = mg
a(m - 1/2 M) = mg
a = (mg)/(m - 1/2 M)
a = (14.2 kg*9.8 m/s²)/(14.2 kg - 1/2*12.8 kg)
a = 17.84 m/s²

substitute back into equation 1
T = ma - mg
T = 14.2 kg*17.84 m/s² - 14.2 kg*9.8 m/s²
T = 114.17 N

This is wrong the answer should be somewhere around 41

Check your algebra where I pointed out the sign error.

You result that says a = 17.84 m/s² should raise a red flag .

 

[DameLight]

Check your algebra where I pointed out the sign error.

You result that says a = 17.84 m/s² should raise a red flag .

I fixed the sign error but now the result is a = -17.84 instead

 

[Nathanael]

You've set it up so that the bucket accelerates downwards when α is positive. But you've also set it up so that downwards acceleration corresponds to a negative "a". Thus your equation should actually be α=-ar
(Or you can make downwards your positive direction, in which case you have to change the other equation to ma=mg-T, either way works.)

 

[haruspex]

T - mg = ma

Here you must be taking g as positive down but a as positive up.

a = rα

Since a is positive up, this makes a positive α correspond to the bucket being accelerated upwards.

T = 1/2 Mrα

But here a positive α correspond to the bucket accelerating downwards.

You need to lay out clearly what your sign conventions are and check your equations adhere to them.

 

[DameLight]

Here you must be taking g as positive down but a as positive up.

Should I have put a as negative? I didn't assign it a sign since I didn't know the direction.

 

[haruspex]

Should I have put a as negative? I didn't assign it a sign since I didn't know the direction.

You don't need to predict the sign of the answer. Just adopt a convention as to which way is positive, which way is negative, and stick to it. If you define up as positive and get a negative answer then that just means the actual answer is downwards.
To help with the discussion, I'm going to assume we're looking at the cylinder end-on, with the rope hanging from the right hand side.
In the present case, if you define up as positive for a then it would be consistent to define anticlockwise as positive for α.
Now, you don't have to do that, you just need to be careful how you write the equations. So, if we take up as positive for a but clockwise as positive for α then the rotational acceleration is given by a = -rα. But you are less likely to go wrong if you choose your sign conventions consistently.

 

[dean barry]

Try this:
have the bucket at an arbitrary constant speed (say 10 m/s)
calculate the KE of both the bucket and the windlass cylinder at this speed
give the windlass an effective linear mass by comparing the KE's