Question

A clock has a continuously moving second’s hand of 0.1 m length. The average acceleration of the tip of the hand (in units of \[m{s^{ - 2}}\]) is of the order of
A. \[{10^{ - 3}}\]
B. \[{10^{ - 1}}\]
C. \[{10^{ - 2}}\]
D. \[{10^{ - 4}}\]

Answer 1

Hint: In this question, we need to determine the order of the average acceleration of the tip of the hand. For this, we need to use the formulae of angular velocity and angular acceleration. After simplification we get the required result.

Formula used:
The formula for angular frequency is given below.
\[\omega = \dfrac{{2\pi }}{T}\]
Here, \[\omega \] is the angular frequency, \[T\] is the period.
Also, the average acceleration is,
\[\alpha = {\omega ^2}R\]
Here, \[\alpha \] is the average acceleration and \[R\] is the length of a radius.

Complete step by step solution:
We know that the second hand has a length of 0.1 m.
That means \[R = 0.1{\text{ m}}\]
Let us find the angular frequency first.
\[\omega = \dfrac{{2\pi }}{T}\]
Here, \[T = 60\]
\[\omega = \dfrac{{2\pi }}{{60}}\]
By simplifying, we get
\[\omega = 0.105{\text{ rad/sec}}\]
Now, we will find the average acceleration.
\[\alpha = {\omega ^2}R\]
\[\Rightarrow \alpha = {\left( {0.105} \right)^2} \times \left( {0.1} \right)\]
By simplifying, we get
\[\alpha = 1.102 \times {10^{ - 3}}{\text{ m/}}{{\text{s}}^2}\]
Hence, the average acceleration of the tip of the hand (in units of \[m{s^{ - 2}}\]) is of the order of \[{10^{ - 3}}\].

Therefore, the correct option is (A).

Additional information: The angular frequency is defined as the angular displacement of any wave component per unit time or even the rate of change of the waveform's stage whereas the average acceleration is defined as the alteration in velocity over a specific period of time.

Note: Many students make mistakes in writing the formulas of angular frequency and average acceleration. If students make a mistake in finding angular frequency then we will get the wrong average acceleration. So, it is necessary to find all the parameters accurately.