Water is drawn from a well in a bucket tied to the end of a rope whose other end

Question

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 a bucket of water is 20 kg.  Someone cranks the bucket up and then lets 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?

- I need to know how to do it, not just the answer, thank you

Explanation / Answer

Relevant equations
? = I?
? = 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

3. The attempt at a solution

So using the equations:
RT = ? = I?
RT= I?

(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)? r = F

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

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

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