Question

A very long uniform helix is made of thin metal wire. The axis of helix is vertical. A small bead begins to slide down the fixed helix starting from rest. Considering friction between bead and wire of helix to be non-zero, which of the following statements is/ are true as long as bead moves on helix.

• A. The speed of bead keeps on increasing.
• B. The magnitude of frictional force on bead remains constant.
• C. The speed of bead first increases and then remains constant.
• D. The magnitude of frictional force increases and then remains constant.

Answer 1

Answer: (C), (D)

The correct options are
C The speed of bead first increases and then remains constant.
D The magnitude of frictional force increases and then remains constant.

From FBD, we can say that, $N+mgsin\theta =\frac{m{v}^2}{r}$ .....(1)

$mgcos\theta -\mu N=m\frac{dv}{dt}$.....(2)

From (1), we can say that at first when velocity will increase due to its weight, Normal (N) will also increase because mgsin$\theta$ will remain constant, as $\theta$ will remain constant in helical path.

Due to increase in N, $\mu N$ will also increase.

So from (2), we can say that after some time when $\mu N$ will become equal to mgcos $\theta$.

we will get $\frac{dv}{dt}=0$

$⟹v=constant$

The component of weight in the direction of velocity of bead will remain constant.

Also at this moment, the bead will acquire terminal velocity because the frictional force and the gravitational pull along the wire both has become equal.

Also due to constant speed, N will also remain constant (from (1)) due to which $\mu N$ (frictional force) will become constant.

Therefore, option C and D would be the appropriate choices.