A bucket of water weighing 10 kg is rotated in a vertical plane. The radius of rotation is 1 m.

A bucket of water weighing 10 kg is rotated in a vertical plane. The radius of rotation is 1 m. The bucket rotates at a speed of 5 m / s. 1.What is the acceleration of the bucket? 2. What will be the weight of the bucket at the bottom of the trajectory?

m = 10 kg.

R = 1 m.

V = 5 m / s.

g = 10 m / s2.

a -?

P -?

The bucket has a centripetal acceleration a, the value of which is expressed by the formula: a = V² / R, where V is the rotation speed, R is the radius of the circle.

a = (5 m / s) ² / 1 m = 25 m / s2.

When passing the lower point of rotation, 2 forces act on the bucket: gravity m * g directed vertically downward, rope tension force N directed vertically upward.

m * a = m * g + N – 2 Newton’s law in vector form.

For projections onto the vertical axis 2, Newton’s law will take the form: m * a = – m * g + N.

N = m * a + m * g = m * (a + g).

According to Newton’s 3 laws, the force N = R.

P = m * (a + g).

P = 10 kg * (25 m / s2 + 10 m / s2) = 350 N.

Answer: at the lower point of rotation, the weight of the bucket is P = 350 N.