Question

A person lowers a bucket into a well by turning the hand crank, as the drawing illustrates. The crank handle moves with a constant tangential speed of 1.22 m/s on its circular path. The rope holding the bucket unwinds without slipping on the barrel of the crank. Find the linear speed with which the bucket moves down the well.

Answer 1

Answer:

$v = 0.305 m/s$

Explanation:

As we know that diameter of hand crank is given as

$D_1 = 0.4 m$

Also the diameter of the shaft on which the rope is wound is given as

$D_2 = 0.1 m$

now we know that both are moving with same angular speed

so we will have

$\frac{v_1}{D_1} = \frac{v_2}{D_2}$

so we will have

$\frac{1.22}{0.4} = \frac{v}{0.1}$

$v = 0.305 m/s$