Question

The centripetal acceleration of a particle in uniform circular motion is \( 18 \, {ms}^{-2} \). If the radius of the circular path is \( 50 \) cm, the change in velocity of the particle in a time of \( \frac{T}{18} \) is:

• A. \( 9 \, {ms}^{-1} \)
• B. \( 2 \, {ms}^{-1} \)
• C. \( 3 \, {ms}^{-1} \)
• D. \( 6 \, {ms}^{-1} \)

Hint:

For uniform circular motion, the velocity change over a given time is calculated using the angular displacement and the sine function.

Answer 1

The Correct Option is C

Step 1: Determine velocity of the particle
The centripetal acceleration is given by: \[ a_c = \frac{v^2}{r} \] Substituting the given values: \[ 18 = \frac{v^2}{0.50} \] \[ v^2 = 18 \times 0.50 = 9 \] \[ v = 3 { ms}^{-1}. \] Step 2: Change in velocity in time \( \frac{T}{18} \)
The time period \( T \) is given by: \[ T = \frac{2\pi r}{v}. \] Substituting \( r = 0.50 \) m and \( v = 3 \) ms\(^{-1}\): \[ T = \frac{2\pi \times 0.50}{3} = \frac{\pi}{3} { s}. \] The time given in the problem is: \[ t = \frac{T}{18} = \frac{\pi}{3} \times \frac{1}{18} = \frac{\pi}{54} { s}. \] The change in velocity for uniform circular motion over time \( t \) is: \[ \Delta v = v \sin \theta. \] Since in a time \( \frac{T}{18} \), the angular displacement is: \[ \theta = \frac{2\pi}{T} \times t = \frac{2\pi}{\pi/3} \times \frac{\pi}{54} = \frac{6}{1} \times \frac{\pi}{54} = \frac{6\pi}{54} = \frac{\pi}{9}. \] Thus, \[ \Delta v = v \sin \frac{\pi}{9}. \] Approximating \( \sin \frac{\pi}{9} \approx 1/3 \): \[ \Delta v = 3 \times \frac{1}{3} = 3 { ms}^{-1}. \] Step 3: Conclusion
Thus, the correct answer is: \[ 3 \, {ms}^{-1}. \]

Answer 2

The centripetal acceleration \( a_c \) of a particle in uniform circular motion is given as \( 18 \, \text{m/s}^2 \). The radius \( r \) of the circular path is \( 50 \, \text{cm} = 0.5 \, \text{m} \). The centripetal acceleration is related to the velocity \( v \) and the radius by the equation:
\( a_c = \frac{v^2}{r} \).
Substituting the known values:
\( 18 = \frac{v^2}{0.5} \),
\( v^2 = 18 \times 0.5 = 9 \),
\( v = 3 \, \text{m/s} \).
The time given is \( \frac{T}{18} \), where \( T \) is the period. The change in velocity \( \Delta v \) after time \( t = \frac{T}{18} \) is related to the angular displacement. The angular velocity \( \omega \) is \( \omega = \frac{2\pi}{T} \).
In time \( t \), the angular displacement is \( \theta = \omega \times t = \frac{2\pi}{T} \times \frac{T}{18} = \frac{2\pi}{18} \). This means that the particle completes \( \frac{1}{18} \) of the circle, corresponding to a total angle of \( \frac{2\pi}{18} = \frac{\pi}{9} \).
The linear speed \( v \) remains constant in magnitude, but the velocity vector changes direction. The change in velocity \( \Delta v \) is given by \( 2v \sin\left(\frac{\theta}{2}\right) = 2 \times 3 \times \sin\left(\frac{\pi}{18}\right) \approx 2 \times 3 \times \frac{\pi}{18} = 1 \times \pi \approx 3 \, \text{m/s} \) (approximating for small angle sine function).
Hence, the change in velocity of the particle in the given time is \( 3 \, \text{m/s} \).