Animated Solution for Physics - Kinematics: A clock has a continuously moving second's hand of $0.1\text{ m}$ length. The average acceleration of the tip of the hand (in ${\text{ms}}^{-2}$) is of the order of

Imagine a classic wall clock ticking away. Focus your attention on the second's hand. As it sweeps across the clock face, the tip of the hand traces out a perfect circle. This is a classic example of Uniform Circular Motion.

Even though the tip of the hand moves at a constant speed, its direction is continuously changing. In physics, any change in velocity—whether in magnitude or direction—requires an acceleration. For an object moving in a circle at a constant speed, this acceleration is always directed towards the center of the circle. We call this the centripetal acceleration.

The magnitude of centripetal acceleration $a_c$ is given by the elegant formula:

Here, $\omega$ represents the angular velocity (how fast the angle is changing), and $R$ is the radius of the circular path. In our problem, the length of the second's hand acts as the radius, so $R=0.1\text{ m}$.

To find $\omega$, we use the relationship between angular velocity and the time period $T$:

For a second's hand, it takes exactly 60 seconds to complete one full revolution. Therefore, $T=60\text{ s}$.

Let's substitute the time period into our angular velocity equation:

Now, we bring this value back into our master equation for centripetal acceleration:

Squaring $0.105$ gives us approximately $0.011025$. Multiplying this by the radius $0.1$ simply shifts the decimal point one place to the left:

To express this in scientific notation, we write:

The question specifically asks for the order of magnitude of the acceleration. The order of magnitude is the power of 10 that most closely approximates the value. Looking at our final result, the power of 10 is $-3$.

Therefore, the order of magnitude is $10^{-3}$, which perfectly matches option (a).