Consider a circular disc which is rotating about its geometric center with a circular motion such...

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Consider a circular disc which is rotating about its geometric center with a circular motion such that the angular acceleration is a quadratic function in time such that the time derivative of the angular acceleration is proportional to the time. For a particular angular acceleration motion at a time of t₁ = 0.217 second the angular acceleration is a₁ = -0.1564 meter per second squared and at a time of t2 = 0.6431 second the angular acceleration is a₂ = 0.8781 meter per second squared. At a time of t3 = 1.0143 second the angular velocity is w=10.2667 radians per second. Calculate the following at time t4 = 2.1075 second and a point on the disc where the radius is r =0.9631 meter: Question 1) the scalar magnitude of the velocity at time 2.1075 second Question 2) the scalar magnitude of the normal component of the vector of the acceleration at time 2.1075 second Question 3) the scalar magnitude of the tangential component of the vector of the acceleration at time 2.1075 second Answer: Consider a circular disc which is rotating about its geometric center with a circular motion such that the angular acceleration is a quadratic function in time such that the time derivative of the angular acceleration is proportional to the time. For a particular angular acceleration motion at a time of t₁ = 0.217 second the angular acceleration is a₁ = -0.1564 meter per second squared and at a time of t2 = 0.6431 second the angular acceleration is a₂ = 0.8781 meter per second squared. At a time of t3 = 1.0143 second the angular velocity is w=10.2667 radians per second. Calculate the following at time t4 = 2.1075 second and a point on the disc where the radius is r =0.9631 meter: Question 1) the scalar magnitude of the velocity at time 2.1075 second Question 2) the scalar magnitude of the normal component of the vector of the acceleration at time 2.1075 second Question 3) the scalar magnitude of the tangential component of the vector of the acceleration at time 2.1075 second Answer:

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To solve the given problem well need to find the angular velocity angular acceleration and position of the point on the disc at time t 21075 seconds Then we can calculate the required quantities Given ... View the full answer