A 49.1-cm diameter disk rotates with aconstant angular acceleration of 2.5rad/s

Question

A 49.1-cm diameter disk rotates with aconstant angular acceleration of 2.5rad/s2. It starts from rest at t = 0, and aline drawn from the center of the disk to a point P on therim of the disk makes an angle of 57.3° with the positivex-axis at this time. (a) Find the angular speed of the wheel att = 2.30 s.
1 rad/s

(b) Find the linear velocity and tangential acceleration of P att = 2.30 s.
linear velocity 2 m/s tangential acceleration 3 m/s2
c) Find the position of P (in degrees, with respect to the positivex-axis) at t = 2.30s.
4° (a) Find the angular speed of the wheel att = 2.30 s.
1 rad/s

(b) Find the linear velocity and tangential acceleration of P att = 2.30 s.
linear velocity 2 m/s tangential acceleration 3 m/s2
c) Find the position of P (in degrees, with respect to the positivex-axis) at t = 2.30s.
4° linear velocity 2 m/s tangential acceleration 3 m/s2

Explanation / Answer

Diameter of the disk = 49.1 cm Radius (r) of the disk = (49.1cm)/2                                 = 24.55 cm                                 = 0.2455 m Angular acceleration () of the disk = 2.5 rad/s2 Initial angular speed (o) of the disk =0 a) We know that = o + t                          = (2.5 rad/s2)(2.30s) =5.75 rad/s b) Linear speed is v = r                           = (0.2455m)(5.75rad/s) = 1.411625 m/s Tangential acceleration is a = r                                          = (0.2455m)(2.5 rad/s2) =0.61375m/s2 c) Position of the diak at t = 2.30s is            = o + (1/2)t2              = 57.3° + (1/2)(2.5rad/s2)(2.30s)2              = 57.3° + (1/2)(2.5rad/s2)(2.30s)2

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