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a flywheel with a radius of 0.3 m starts from rest and accelerates with a constant angular acceleration of 0.6 rad/s². compute the magnitude of the tangential acceleration, the centripetal acceleration, and the resultant acceleration of a point on its rim after it has turned throuh 1 rad

User IGian
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Final answer:

The magnitude of the tangential acceleration, centripetal acceleration, and resultant acceleration of a point on the rim of a flywheel are 0.18 m/s², 0.365 m/s², and 0.408 m/s².

Step-by-step explanation:

The magnitude of the tangential acceleration can be calculated using the equation: tangential acceleration = angular acceleration * radius.

In this case, the radius is given as 0.3 m and the angular acceleration is given as 0.6 rad/s².

So, the tangential acceleration is 0.6 * 0.3

= 0.18 m/s².

The centripetal acceleration can be calculated using the equation: centripetal acceleration = (angular velocity)² * radius.

After the flywheel has turned through 1 rad, its angular velocity can be calculated using the equation: angular velocity = sqrt(2 * angular acceleration * angle).

In this case, the angle is given as 1 rad and the angular acceleration is given as 0.6 rad/s².

So, the angular velocity is sqrt(2 * 0.6 * 1) = 1.095 rad/s.

Substituting this value and the given radius of 0.3 m into the centripetal acceleration equation, we get:

centripetal acceleration = (1.095)² * 0.3

= 0.365 m/s².

The resultant acceleration at a point on the rim of the flywheel can be calculated using the formula:

resultant acceleration = sqrt(tangential acceleration² + centripetal acceleration²).

Substituting the previously calculated values, we get: resultant acceleration

= sqrt(0.18² + 0.365²)

= 0.408 m/s².

User Jobeso
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