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The angular position of a point on a rotating wheel is given by θ = 5 3t2 2t3, where θ is in radians and t is in seconds. (a) at t = 0, what is the point's angular position? rad (b) at t = 0, what is the point's angular velocity? rad/s (c) what is the point's angular velocity at t = 4.0 s? rad/s (d) calculate the point's angular acceleration at t = 2.0 s. rad/s2 (e) is the angular acceleration constant? yes no not enough information

User Ross Riley
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Final answer:

The point's angular position at t=0 is 0 radians, and its angular velocity at t=0 is also 0 rad/s. The point's angular velocity at t=4s and its angular acceleration at t=2s can be calculated using the first and second derivatives of the angular position, respectively. The angular acceleration is not constant, as it changes with time.

Step-by-step explanation:

The angular position given for a rotating wheel is θ = 5t3 - 2t3, where θ is in radians and t is in seconds. This information allows us to solve for various properties of rotational motion such as angular position, angular velocity, and angular acceleration.

For t = 0:

  • Angular Position: Plugging t = 0 into the equation, we get θ = 0 radians.
  • Angular Velocity: The first derivative of θ with respect to time (t) gives angular velocity (ω = dθ/dt). At t = 0, the angular velocity is 0 rad/s since the derivative of the constant and cubic term are zero when t = 0.

At t = 4 seconds:

  • The angular velocity is found by taking the first derivative of θ with respect to t, which gives ω = 15t2 - 6t2, and evaluating it at t = 4 s.

At t = 2 seconds:

  • The angular acceleration (α) is the second derivative of θ with respect to t, which can be calculated and evaluated at t = 2 s.

Angular Acceleration Constancy:

  • Angular acceleration is not constant as it depends on time t and is given by the second derivative of θ (α = d2θ/dt2).

User WilHall
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