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? A)1.1 $\times$ 10² rad/s² B)3.6 rad/s² C)35 rad/s² D)63 rad/s² E)17 rad/s²

Question

Quizplus · Accepted Answer

The force on the cylinder due to the tension in the rope is given by:

F = (m_cylinder + m_bucket)g

where m_cylinder is the mass of the cylinder (50 kg), m_bucket is the mass of the water in the bucket (20 kg), and g is the acceleration due to gravity (9.8 m/s^2).

The torque on the cylinder due to this force is given by:

τ = Fr

where r is the radius of the cylinder (0.125 m).

The moment of inertia of a solid cylinder about its axis of rotation is given by:

I = (1/2)mr^2

where m is the mass of the cylinder (50 kg).

The angular acceleration of the cylinder is given by:

α = τ/I

Substituting the above expressions, we get:

α = (2(m_cylinder + m_bucket)g)/(3mr)

Plugging in the values, we get:

α = (2(50 + 20)9.8)/(3*50*0.125) = 35 rad/s^2

Therefore, the answer is C.

Water Is Drawn from a Well in a Bucket Tied ×\times×

Water Is Drawn from a Well in a Bucket Tied $\times$