A flywheel with a radius of 0.390 m starts from rest and accelerates with a cons

Question

A flywheel with a radius of 0.390 m starts from rest and accelerates with a constant angular acceleration of 0.640 rad/s2 .

Part A

Compute the magnitude of the tangential acceleration, the radial acceleration, and the resultant acceleration of a point on its rim at the start.

Enter your answers separated with commas.

Part B

Compute the magnitude of the tangential acceleration, the radial acceleration, and the resultant acceleration of a point on its rim after it has turned through 60.0 .

Enter your answers separated with commas.

Part C

Compute the magnitude of the tangential acceleration, the radial acceleration, and the resultant acceleration of a point on its rim after it has turned through 120 .

Enter your answers separated with commas.

Explanation / Answer

Given

   radius of the flywheel is r = 0.390 m

starts from rest and the constant angular acceleration is 0.640 rad/s2

partA

   tangential acceleration is a_t = r*alpha = 0.39*0.640 = 0.2496 m/s2

   radial acceleration is a_r = v^2/r = 0 m/s2 , v=0 m/s

and the tangential acceleration is cosntant it does not change

so the magnitude of the acceleration of the fly wheel is a = sqrt(a_t^2+ a_r^2) = sqrt(0.2496^2+0^2) = 0.2496 m/s2

PArt B

   Tangential acceleration is same a_t = 0.2496 m/s2

radial acceleration when the fly wheel is turned through 60 degrees is

using equations of motion v^2-u^2 = 2as

           W2^2 - W1^2 = 2 alph*theta

       W1=0 ==> W2 = sqrt(2*alpha*theta)
           W2 = sqrt(2*0.640*pi/3) m/s2

           W2 = 1.157762 m/s2
the magnitude of the acceleration of the fly wheel is a = sqrt(a_t^2+ a_r^2) = sqrt(0.2496^2+1.157762^2) m/s2 = 1.184362 m/s2

Part C

   when the wheel turned through 120 degrees = 2pi/3 radians

   Tangential acceleration is same a_t = 0.2496 m/s2

radial acceleration when the fly wheel is turned through 120 degrees is

using equations of motion v^2-u^2 = 2as

           W2^2 - W1^2 = 2 alph*theta

       W1=0 ==> W2 = sqrt(2*alpha*theta)
           W2 = sqrt(2*0.640*2pi/3) m/s2

           W2 = 1.63732 m/s2

the magnitude of the acceleration of the fly wheel is a = sqrt(a_t^2+ a_r^2) = sqrt(0.2496^2+1.63732^2) m/s2 = 1.656236 m/s2

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