Numerical Problems Based on Uniform Circular Motion for Class 11 Physics

Angular velocity is given by $\omega = \frac{2\pi}{T}$, where $T$ is the time period of one full revolution.

(i) For the second hand:
It completes one revolution in $60\text{ seconds}$ ($T = 60\text{ s}$).

(ii) For the minute hand:
It completes one revolution in $60\text{ minutes}$ ($T = 60 \times 60 = 3600\text{ s}$).

Given: mass $m = 0.4\text{ kg}$, radius $r = 2\text{ m}$, linear speed $v = 10\text{ ms}^{-1}$.

(i) Angular speed ($\omega$):

(ii) Frequency of revolution ($f$):

(iii) Time period ($T$):

(iv) Centripetal acceleration ($a_c$):

Given: Radius $r = 0.50\text{ m}$, Linear speed $v = 10\text{ ms}^{-1}$.

The relationship between linear speed and angular speed is $v = r\omega$. Solving for $\omega$:

First, convert the given units into standard SI units (meters and seconds).

Radius $R = 3.85 \times 10^5\text{ km} = 3.85 \times 10^8\text{ m}$.
Time period $T = 27.3\text{ days} = 27.3 \times 24 \times 60 \times 60\text{ s} = 2,358,720\text{ s}$.

The angular velocity ($\omega$) is:

The acceleration of the moon towards the earth is its centripetal acceleration ($a_c$):

Given: Radius $r = 50\text{ cm}$, Time $t = 5\text{ min} = 300\text{ s}$.
Initial frequency $f_1 = 100\text{ rpm}$, Final frequency $f_2 = 400\text{ rpm}$.

Convert frequencies to initial and final angular velocities (rad/s):

(i) Angular acceleration ($\alpha$):

(ii) Linear (tangential) acceleration ($a_t$):

Using the radius in cm to match the given answer unit ($\text{cm s}^{-2}$):