A 40.2-cm diameter disk rotates with aconstant angular acceleration of 2.3rad/s

Question

A 40.2-cm diameter disk rotates with aconstant angular acceleration of 2.3rad/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 positive x-axis) att = 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 positive x-axis) att = 2.30s.
4° linear velocity 2 m/s tangential acceleration 3 m/s2

Explanation / Answer

   Given  angularacceleration      =   2.3   rad/s2             radiusofdisk   r   =   d/2                                              =   0.402/ 2                                              =   0.201   m          Initialangle   0   =   57.30                                        =   57.3* / 180                                        =   0.9996      rad                                        =   57.3* / 180                                        =   0.9996      rad          Initialangularspeed   0   =   0    a.   first equation forangular motionis            =   0   +   * t                                                                                     =   0   +   2.3* 2.30                                                                                     =   5.29   rad/s    b.   Linearvelocity      v   =   r*                                              =   0.201* 5.29                                              =   1.06   m/s          tangentialacceleration   a   =   r*                                                          =   0.201* 2.30                                                          =   0.462   m/s2    c.   Angular position         =   0   +   0* t   +   (1/2) * *t2                                                 =   0.9996   +   0* 2.3   +   0.5 * 2.3 *2.302                                                 =   7.083   rad                                                 =   7.083* 180 /                                                 =   406.040          Sinceafter 3600, positions will be repeated,          ( from + ve x axis)   =   406.04 -360                                                 =   46.040          Sinceafter 3600, positions will be repeated,          ( from + ve x axis)   =   406.04 -360                                                 =   46.040

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