Question

A block of 200 g mass moves with a uniform speed in a horizontal circular groove, with vertical sidewalls of a radius of 20 cm. If the block takes 40 s to complete one round, the normal force by the sidewalls of the groove is:
A. \[6.28 \times {10^{ - 3}}{\text{ N}}\]
B. \[0.0314{\text{ N}}\]
C. \[9.859 \times {10^{ - 2}}{\text{ N}}\]
D. \[9.859 \times {10^{ - 4}}{\text{ N}}\]

Answer 1

Hint: In this question, we need to find the normal force by the sidewalls of the groove. For this, we have to use the formula of normal force for finding the force exerted by the sidewalls of the groove.

Formula used:
We will use the following formula of force.
\[F = m{\omega ^2}R\]
Here, \[F\] is a normal force, \[m\] is the mass of an object, \[\omega \] is the angular frequency and \[R\] is the radius.

Complete step by step solution:
We know that the normal force will offer the essential centripetal force.
So, the normal force can be calculated as
\[F = m{\omega ^2}R\]
Let us first convert all the parameters into SI units.
We know that \[1\text{ kg}=1000\text{ g}\]
Here, \[m = 200g \Rightarrow \dfrac{{200}}{{1000}} = 0.2{\text{ kg}}\]
Also, \[1\text{ m}=100\text{ cm}\]
So, \[R = 20cm \Rightarrow \dfrac{{20}}{{100}} = 0.2{\text{ m}}\]
Let us find the angular frequency.
\[\omega = \dfrac{{2\pi }}{T} \Rightarrow \dfrac{{2\pi }}{{40}} = \dfrac{\pi }{{20}}\]
So, we get
\[F = \left( {0.2} \right) \times {\left( {\dfrac{\pi }{{20}}} \right)^2} \times 0.2\]
\[\Rightarrow F = \left( {0.2} \right) \times \left( {\dfrac{{3.14 \times 3.14}}{{20}}} \right) \times 0.2\]
By simplifying further, we get
\[\therefore F = 9.859 \times {10^{ - 4}}{\text{ N}}\]
Hence, we can say that the normal force by the sidewalls of the groove is \[9.859 \times {10^{ - 4}}{\text{ N}}\]

Therefore, the correct option is (D).
Note:The normal force is the force exerted by surfaces to avoid solid objects from passing through them. A contact force is a normal force. A normal force cannot be exerted on two surfaces that are not in interaction.