A circular saw blade 0.200 m in diameter starts from rest. In 6.00 s it accelerates with constant angular acceleration to an angular velocity of 140 rad/s. Find the angular acceleration and the angle through which the blade has turned.

Short Answer

Expert verified

The angular acceleration, and angle are 23.33rad/s2, and 420 rad, respectively.

Step by step solution

01

Identification of given data

The blade starts from the rest, so the initial angular speedis ω0z=0rad/s.

The final angular speed is ωz=140rad/s.

The time is t=6.00s.

02

Concept/Significance of rotation of body with constant angular acceleration

Angular velocity at time t of a rigid body with constant angular acceleration is given by,

ωz=ω0z+αzt........(1)

Here,αz is the angular acceleration,ω0z is the angular velocity of body at time 0, t is the time.

Angular position at time t of a rigid body with constant angular acceleration is given by,

θ=ω0zt+12αzt2........(2)

03

Determine the angular acceleration and the angle through which the blade has turned

Substitute, ω0z=0rad/s,t=6s andωz=140rad/s in equation (1).

140rad/s=06s+αz6sαz=140rad/s6s=23.33rad/s2

Therefore, the required angular accelerationis23.33rad/s2 .

Substitute θ0=0rad, t=6s, ω0z=0rad/s, andαz=23.33rad/s2 in equation (2).

θ=0+06s+1223.33rad/s226s420rad

Therefore, the angle is 420 rad.

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