A flywheel with a radius of 0.600 m starts from rest and accelerates with a cons

Question

A flywheel with a radius of 0.600 m starts from rest and accelerates with a constant angular acceleration of 0.500 rad/s2 .

1-Compute the magnitude of the tangential acceleration of a point on its rim at the start.

2-Compute the magnitude of the radial acceleration of a point on its rim at the start.

3-Compute the magnitude of the tangential acceleration of a point on its rim after it has turned through 60.0

4-Compute the magnitude of the radial acceleration of a point on its rim after it has turned through 60.0 .

5-Compute the magnitude of the tangential acceleration of a point on its rim after it has turned through 120.0 .

6-Compute the magnitude of the radial acceleration of a point on its rim after it has turned through 120.0 .

Explanation / Answer

1) The tangential acceleration is the angular acceleration multiplied by the radius:

At = alpha*r = 0.500*0.600 = 0.3 m/sec^2

2.)The radial since the velocity is 0.
3.)The radial centripetal acceleration is 0 as velocity is 0.

The resultant magnitude is the tangential acceleration:

A = At = 0.18m/sec^2

B) The angle is (1/2)*alpha*t^2, solving for t:

t = sqrt(2*angle/alpha) = sqrt[(2*60*pi/180)/.5] = 2.0466 sec

omega = alpha*t = .5*2.0466 = 1.0233 rad/sec

The centripetal (or radial) acceleration is:

Ar = omega^2*r = 1.0233^2*.6 = 0.628 m/sec^2.

Since the angular acceleration is constant, the tangential acceleration is still:

At = alpha*r = 0.3m/sec^2

Since the components are at 90 degrees to each other, you can find the magnitude of the resultant by applying the Pythagorean Theorem:

A = sqrt(At^2 + Ar^2) = sqrt(.3^2 + .0.628^2) = .6959m/sec^2

4)
Ar = omega^2*r = 1.0233^2*.6 = 0.628 m/sec^2

5.)

The tangential acceleration is still as computed above:

At = .3m/sec^2




6)

t = sqrt(2*angle/alpha) = sqrt[(2*120*pi/180)/.5] = 2.89 sec

omega = alpha*t = .5*2.89 = 1.445 rad/sec

The centripetal (or radial) acceleration is:
Ar = omega^2*r = 1.445^2*.6 = 1.252m/sec^2.

The magnitude of the resultant is:

Ar =sqrt(.3^2 + 1.252^2) = 1.287 m/sec^2

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