Angular Displacement Differentiation

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SUMMARY

The discussion focuses on calculating angular displacement, velocity, and acceleration using the equation θ = sin(3t). The angular velocity at t = 1 second is determined to be approximately -2.97 rad/s, while the angular acceleration at t = 0.5 seconds is calculated as -8.98 rad/s². For parts b and d, the smallest positive values of t for which the angular velocity is 2 rad/s and the angular acceleration is 9 rad/s² are found to be 0.280 seconds and 1.57 seconds, respectively. The method involves using the derivative of the sine function and applying inverse cosine for solving the equations.

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  • Understanding of angular displacement and its mathematical representation
  • Knowledge of differentiation, specifically d/dx (sin ax) = a cos ax
  • Familiarity with angular velocity and angular acceleration concepts
  • Ability to use inverse trigonometric functions for solving equations
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  • Study the application of derivatives in physics, particularly in rotational motion
  • Learn about inverse trigonometric functions and their applications in solving equations
  • Explore the relationship between angular displacement, velocity, and acceleration in detail
  • Practice additional problems involving angular motion and differentiation techniques
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Homework Statement


An angular displacement θ radians in time t seconds is given by the equation θ = sin 3t. Find

a:) angular velocity when t = 1 second
b:) the smallest positive value of t for which the angular velocity is 2rad/s
c:) the angular acceleration when t = 0.5 seconds
d:) the smallest positive value of t for which the angular acceleration is 9rad/s


Homework Equations


d/dx (sin ax) = a cos ax
d/dx (cos ax) = -a sin ax


The Attempt at a Solution


Ok i have worked out parts a:) and c:)

a:) dθ/dt = 3cos 3t so when t = 1, ω = -2.97r/s

c:) dω/dt = -9sin 3t so when t = 0.5, angular acceleration = -8.98r/s/s

I can't work out parts b and d, its not a homework question i am revising from a book and the answers for parts b and d are 0.280seconds and 1.57 seconds respectively however the method is not explained. I managed to calculate the answer to part d at one point but cannot remember how i got there :shy:

Any help would be appreciated, thanks
 
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This really just boils down to saying, for example on part (b),

\dot \theta = 3 \cos 3t = 2

Move the 3 over so you get \cos 3t = 2/3 and then use an inverse cosine function. That's really all there is to it; (d) is very similar.
 
Ok that's great thanks!
 

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