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Rotational motion with constant angular acceleration (1) When a carpenter shuts off his circular saw, the 10.0 inch diameter blade slows from 3984 revolutions per minute to rest in 1.25 s. (a) How many revolutions does a point on the rim of the blade rotate through during the deceleration? (answer: 41.5 revolutions) (b) What angle does a point on the rim of the blade rotate through during the deceleration? (answer: 261 radians) (c) What is the distance traveled by a point on the rim of the blade during the deceleration? (answer: 109 feet) (2) A Ferris wheel with radius 14.8 m is speeding up. At a particular instant, the wheel has an angular speed of 0.18 rad/s and an angular acceleration of 0.28 rad/s2. Calculate · the magnitude of the total acceleration of a point on the edge of the wheel. (answer: 4.2 m/s2) · the direction of the total acceleration with respect to the tangential direction of motion. (answer: 6.6 degrees) (3) A dentist’s polishing wheel goes from an angular speed of 0 rad/s to 640.1 rad/s in 2.76 s with constant angular acceleration. What are the magnitude and direction (relative to the tangential direction) of the acceleration of a point on the rim of the wheel 0.086 s after it starts spinning? The radius of the wheel is 3.20 mm. (answer: magnitude = 1.47 m/s2; direction = 59.8 degees).

Definition Definition Rate of change of angular velocity. Angular acceleration indicates how fast the angular velocity changes over time. It is a vector quantity and has both magnitude and direction. Magnitude is represented by the length of the vector and direction is represented by the right-hand thumb rule. An angular acceleration vector will be always perpendicular to the plane of rotation. Angular acceleration is generally denoted by the Greek letter α and its SI unit is rad/s 2 .

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Rotational Motion with constant angular acceleration:

The motion of a point on the body rotating with constant angular acceleration about an axis is defined by the following three equations,

$\begin{array}{rcl} ω & = & ω_{0} + α t (1) \\ ∆ θ & = & ω_{0} t + \frac{1}{2} α t^{2} (2) \\ ω^{2} & = & ω_{0}^{2} + 2 α ∆ θ (3) \end{array}$

where $ω$ is the final angular velocity, $ω_{0}$ is the initial angular velocity, $∆ θ$ is the angular displacement, $α$ is the angular acceleration, and $t$ is the time.

The distance traveled by the point at $r$ distance from the axis of rotation is given by the equation,

$d = r ∆ θ (4)$

NOTE:

$1 Revolution per minute (or rpm)= \frac{1 Revolution per second (or rps)}{60} 1 inch = 0.083333 feet$

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