Given:
- Radius of the circular track: \( R = 9 \, \text{m} \)
- Number of revolutions completed: 120 revolutions
- Time taken: 3 minutes
Step 1: Calculate the Angular Velocity
The angular velocity \( \omega \) is given by:
\[ \omega = \frac{\text{Total revolutions}}{\text{Time taken}} \times 2\pi \, \text{rad/s}. \]
Substituting the given values:
\[ \omega = \frac{120 \, \text{revolutions}}{3 \, \text{minutes}} \times \frac{2\pi \, \text{rad}}{1 \, \text{revolution}}. \]
Converting time to seconds:
\[ \omega = \frac{120 \times 2\pi}{3 \times 60} \, \text{rad/s} = \frac{4\pi}{3} \, \text{rad/s}. \]
Step 2: Calculate the Centripetal Acceleration
The centripetal acceleration \( a_{\text{centripetal}} \) is given by:
\[ a_{\text{centripetal}} = \omega^2 R. \]
Substituting the values of \( \omega \) and \( R \):
\[ a_{\text{centripetal}} = \left(\frac{4\pi}{3}\right)^2 \times 9. \]
Simplifying:
\[ a_{\text{centripetal}} = \frac{16\pi^2}{9} \times 9 = 16\pi^2 \, \text{m/s}^2. \]
Therefore, the magnitude of the centripetal acceleration of the monkey is \( 16\pi^2 \, \text{m/s}^2 \).
A body of mass 100 g is moving in a circular path of radius 2 m on a vertical plane as shown in the figure. The velocity of the body at point A is 10 m/s. The ratio of its kinetic energies at point B and C is: (Take acceleration due to gravity as 10 m/s^2)
Let \( y = f(x) \) be the solution of the differential equation
\[ \frac{dy}{dx} + 3y \tan^2 x + 3y = \sec^2 x \]
such that \( f(0) = \frac{e^3}{3} + 1 \), then \( f\left( \frac{\pi}{4} \right) \) is equal to:
Find the IUPAC name of the compound.
If \( \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p \), then \( 96 \ln p \) is: 32