Question

A small block of mass 100 g moves with uniform speed in a horizontal circular groove. The radius of the groove’s vertical sidewalls is 25 cm. If the block takes 2 s to complete one round, find the normal contact force by the sidewall of the groove.
A) 0.25N
B) 1.05N
C) 0.75N
D) 0.5N

Answer 1

Hint: The block constitutes uniform circular motion as it moves along the groove with a uniform speed. Its centripetal force must balance the normal contact force by the sidewalls so that the block can keep moving.

Formula Used:
1) The centripetal force acting on a body of mass $m$ moving with a velocity $v$ along a circle of radius $r$ is given by, ${F_c} = \dfrac{{m{v^2}}}{r}$
2) The speed of a body is given by, $v = \dfrac{d}{t}$ where $d$ is the distance covered and $t$ is the time taken by the block to cover the distance.

Complete step by step answer:
Step 1: Write down the parameters given in the question.
The mass of the block moving along the circular groove is $m = 100{\text{g}} = {\text{0}}{\text{.1kg}}$ .
The time taken by the block to complete a single round is $t = 2{\text{s}}$ .
Also given that the radius of the vertical sidewall is $r = 25{\text{cm}} = 0.25{\text{m}}$ .
Step 2: Find the velocity of the block.
Let $v$ be the uniform speed of the block.
We have the expression for speed as $v = \dfrac{d}{t}$ where $d$ is the distance covered and $t$ is the time taken by the block.
The distance covered will be $d = 2\pi r = 2\pi \times 0.25$
Substituting the values for $t = 2{\text{s}}$ and $d = 2\pi \times 0.25$ in the above relation we get, $v = \dfrac{{2\pi \times .25}}{2} = 0.25\pi {\text{ m/s}}$
Step 3: Find the normal contact force.
The centripetal force ${F_c}$ of the block will be equal to the normal force $N$ applied by the sidewall of the groove i.e., ${F_c} = N$ .
The relation for the centripetal force is given by, ${F_c} = \dfrac{{m{v^2}}}{r}$ ------- (1)
where $m$ is the mass of the body, $v$ is its uniform speed and $r$ is the radius of the sidewall of the groove.
Substituting values for $m = {\text{0}}{\text{.1kg}}$ , $r = 0.25{\text{m}}$ and $v = 0.25\pi {\text{ m/s}}$ in equation (1) we get, ${F_c} = \dfrac{{0.1 \times {{\left( {0.25\pi } \right)}^2}}}{{0.25}} = 0.25{\text{N}}$
$\therefore$ the normal contact force by the sidewall of the groove is $N = 0.25{\text{N}}$ .
$\therefore$ The correct option is (A)

Note:
The block is said to move along a horizontal circular groove. This implies that it has a circular motion. Thus the distance covered in one round will be the circumference of the groove. Hence we calculate the distance as $d = 2\pi r$ . When substituting values in equations all the physical quantities involved in the equation must be represented in their respective S. I. units.