Angular acceleration of a bucket attatched to a solid cylinder

[anomalocaris]

Homework Statement

This was a question from my homework. I got it wrong, even after asking my professor about it, and even though I can't get credit for it now, I'd like to know where I went wrong if anyone can help sort me out!

"Water is drawn from a well in a bucket tied to the end of a rope whose other end wraps around a solid cylinder of mass 50 kg and diameter 25 cm. As this cylinder is turned with a crank, the rope raises the bucket. The mass of the bucket of water is 20 kg. Someone cranks the bucket up and then let's go of the crank, and the bucket of water falls down to the bottom of the well. Without friction or air resistance, what is the angular acceleration of the 50-kg cylinder?"

m_c (mass of cylinder)=50 kg
d=25 cm so r=12.5 cm = 0.125 m
m_b (mass of bucket)=20 kg

Homework Equations

\tau = I\alpha
\tau = rT
T-(m_b)g=-(m_b)a

Also I'm assuming since it's a solid cylinder, we will need I = 1/2 mr^2

The Attempt at a Solution

So using the equations:
RT = \tau = I\alpha
RT= I\alpha

(m_b)g-T= (m_b)aR And from what I understand, this is the same as the tangential acceleration?

(m_b)g-T=(m_b)\alpha r = F

T= ( i\alpha ) / r
(m_b)g -(( i\alpha ) / r ) = m\alpha r
\alpha ( ((m_b)r) + (I /R ) ) = (m_b)g

Leaving us with the final : \alpha = ((m_b)g)/(((m_b)r) + (I /r))

Using this equation, I found I = 0.390625
and the final answer would be 35 rad/s^2

Sorry for such a long post--this is my first time on the website and I read the rules so hopefully I've done everything correctly! Thank you all!

 

[ehild]

Hi anomalocaris, welcome to PF.

Homework Statement

This was a question from my homework. I got it wrong, even after asking my professor about it, and even though I can't get credit for it now, I'd like to know where I went wrong if anyone can help sort me out!

"Water is drawn from a well in a bucket tied to the end of a rope whose other end wraps around a solid cylinder of mass 50 kg and diameter 25 cm. As this cylinder is turned with a crank, the rope raises the bucket. The mass of the bucket of water is 20 kg. Someone cranks the bucket up and then let's go of the crank, and the bucket of water falls down to the bottom of the well. Without friction or air resistance, what is the angular acceleration of the 50-kg cylinder?"

m_c (mass of cylinder)=50 kg
d=25 cm so r=12.5 cm = 0.125 m
m_b (mass of bucket)=20 kg

Homework Equations

\tau = I\alpha
\tau = rT
T-(m_b)g=-(m_b)a

Also I'm assuming since it's a solid cylinder, we will need I = 1/2 mr^2

The Attempt at a Solution

So using the equations:
RT = \tau = I\alpha
RT= I\alpha

(m_b)g-T= (m_b)aR And from what I understand, this is the same as the tangential acceleration?

a is equal to the tangential acceleration of the rim of the cylinder, and at the same time, the downward acceleration of the bucket. α (alpha) is the angular acceleration of the cylinder. alpha=a/r. So the correct equation is

m_b g-T=m_b\alpha r

T= ( i\alpha ) / r.

(m_b)g -(( i\alpha ) / r ) = m_b\alpha r
\alpha ( ((m_b)r) + (I /R ) ) = (m_b)g

Leaving us with the final : \alpha = ((m_b)g)/(((m_b)r) + (I /r))

Using this equation, I found I = 0.390625
and the final answer would be 35 rad/s^2

Sorry for such a long post--this is my first time on the website and I read the rules so hopefully I've done everything correctly! Thank you all!

Do not use different notations for the same thing: r,R and i,I.

Your final equation

\alpha=\frac{m_b g}{m_b r+I/r}

is correct.
Notice that I=0.5 m_cr², so you can rewrite the equation as

\alpha=\frac{m_b g}{r(m_b +0.5m_c)}
The numerical result for the angular acceleration is also correct.

ehild

 

[anomalocaris]

Thank you, ehild! Sorry for the different notations--that was kind of a typographical error on my part. I'll have to ask my professor why the computer system scored me wrong--maybe there was a glitch somewhere. I really appreciate your help!