Question

A student has 67-cm-long arms. What is the minimum angular velocity (in rpm) for swinging a bucket of water in a vertical circle without spilling any? The distance from the handle to the bottom of the bucket is 35 cm .

Answer 1

Answer:

29.61 rpm.

Explanation:

Given,

student arm length, l = 67 cm

distance of the bucket, r = 35 m

Minimum angular speed of the bucket so, the water not fall can be calculated by equating centrifugal force with weight.

Now,

$mg = m r \omega^2$

$\omega = \sqrt{\dfrac{g}{R}}$

R = 67 + 35 = 102 cm = 1.02 m

$\omega = \sqrt{\dfrac{9.81}{1.02}}$

$\omega = 3.101\ rad/s$

$\omega = \dfrac{3.101}{2\pi} = 0.494\ rev/s$

$\omega = 0.494 \times 60 = 29.61\ rpm$

minimum angular velocity is equal to 29.61 rpm.

Answer 2

Answer:

29.6 rpm

Explanation:

length of arm = 67 cm

distance of handle to the bottom = 35 cm

radius of rotation, R = 67 + 35 = 102 cm = 1.02 m

The centripetal force acting on the bucket is balanced by the weight of the bucket.

mRω² = mg

R x ω² = g

$\omega = \sqrt\frac{g}{R}$

$\omega = \sqrt\frac{9.8}{1.02}$

ω = 3.1 rad/s

Let f is the frequency in rps

ω = 2 x 3.14 x f

3.1 = 2 x 3.14 xf

f = 0.495 rps

f = 29.6 rpm