A 41.7-cm diameter disk rotateswith a constant angular acceleration of 2.9 rad/s

Question

A 41.7-cm diameter disk rotateswith a constant angular acceleration of 2.9 rad/s2. It starts from restat t = 0, and a line drawn from the center ofthe disk to a point P on the rim of the diskmakes an angle of 57.3° with the positive x-axisat this time. (a) Find the angular speed of the wheel at t =2.30 s.
rad/s

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

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

Explanation / Answer

Given      radius   =   d /2   =   41.7 /2   =   0.2085 m

initial angularposition   0   =   57.30,  

angularacceleration      =   2.9 rad/s2

time   t   =   2.30   s,

initial angularvelocity   0   =   0

    a.      Finalangularvelocity      =   0+ * t

             2.3   =   0+ 2.9 * 2.3

             2.3   =  8.7 rad/s

    b.   Linearvelocity   v   =   r *   =   0.2085 * 8.7

          v   =   1.81395 m/s

          tangentialacceleration   a   =   r*    =   0.2085 * 2.9

          a   =  0.6045 m/s2

    c.   angulardisplacement      =   0*t   +   (1/2) * *t2

             =   0* 2.3 + 0.5 * 2.9 * 2.32   =  12.1945 rad.

          1   rad   =   1800/   =>      =   180* 12.1945 / 3.14   =  699.050

          Finalposition of point P   =    +0   699.05 +57.3   =   756.340

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