Answer:
The magnitude of the acceleration of the tip of the minute hand of the clock [tex]1.675\times10^{-6}\ m/s^2[/tex].
Explanation:
Given that,
Length of minute hand = 0.55 m
Length of hour hand = 0.26 m
The time taken by the minute hand to complete one revelation is
[tex]T= 3600\ sec[/tex]
We need to calculate the angular frequency
Using formula of angular frequency
[tex]\omega=\dfrac{2\pi}{T}[/tex]
Put the value into the formula
[tex]\omega=\dfrac{2\pi}{3600}[/tex]
[tex]\omega=0.001745\ rad/s[/tex]
We need to calculate the magnitude of the acceleration of the tip of the minute hand of the clock
Using formula of acceleration
[tex]a=r\omega^2[/tex]
Put the value into the formula
[tex]a=0.55\times(0.001745)^2[/tex]
[tex]a=1.675\times10^{-6}\ m/s^2[/tex]
Hence, The magnitude of the acceleration of the tip of the minute hand of the clock [tex]1.675\times10^{-6}\ m/s^2[/tex].