A flywheel with a radius of 0.240 m starts from rest and accelerates with a cons

Question

A flywheel with a radius of 0.240m starts from rest and accelerates with a constant angular acceleration of

0.662 rad/s2

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

Ans is 0.158,0,0.158     m/s^2

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 degrees

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

Explanation / Answer

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

At = alpha*r = 0.662*0.240 = 0.158m/sec^2

The radial, or centripetal acceleration is 0, since the velocity is 0.

The resultant magnitude is the tangential acceleration:

A = At = 0.158m/sec^2

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

t = sqrt(2*angle/alpha) = sqrt[(2*60*pi/180)/.662] = 1.779 sec

omega = alpha*t = .662*1.779 = 1.177 rad/sec

The centripetal (or radial) acceleration is:

Ac = omega^2*r = 1.177^2*.240 = 0.332m/sec^2.

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

At = alpha*r = .158m/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 + Ac^2) = sqrt(.158^2 + .332^2) = .359m/sec^2

C)

The tangential acceleration is still as computed above:

At = .158m/sec^2

t = sqrt(2*angle/alpha) = sqrt[(2*120*pi/180)/.662] = 2.515 sec

omega = alpha*t = .662*2.515 = 1.664 rad/sec

The centripetal (or radial) acceleration is:

Ac = omega^2*r = 1.664^2*.240 = 0.664m/sec^2.

The magnitude of the resultant is:

sqrt(.158^2 + .664^2) = .682m/sec^2

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