A wheel 2.00 m in diameter lies in a vertical plane androtates with a constant a

Question

A wheel 2.00 m in diameter lies in a vertical plane androtates with a constant angular acceleration of 4.00rad/sec2. The wheel starts at rest at t=0, and theradius vector of point P on the rim makes an angle of 57.3 degreeswith the horizontal at this time. At t=2.00 s, find: a) the angular speed of their wheel b) the linear speed and acceleration of the point P c) the angular position at point P. A wheel 2.00 m in diameter lies in a vertical plane androtates with a constant angular acceleration of 4.00rad/sec2. The wheel starts at rest at t=0, and theradius vector of point P on the rim makes an angle of 57.3 degreeswith the horizontal at this time. At t=2.00 s, find: a) the angular speed of their wheel b) the linear speed and acceleration of the point P c) the angular position at point P.

Explanation / Answer

   Initial angularspeed   0   =   0,   angularaccleration      =   4.00   rad/s2,   time   t   =   2.00s,   radius   r   =   2.00m    Final angularspeed  =    ,   Finalposition   =   ,   Initialposition  0   =   57.30    a.      =   0   +   *t   =   0   +   4.00*2.00   =   8.00   rad/s    b.   Linear speed ofP   v   =   r *   =   2.00 *8.00   =   16.00   m/s          Linearacceleration      a   =   r*    =   2.00 *4.00   =   8.00   m/s2    c.   angulardisplacement         =     (0* t   +   0.5 * *t2)                =   0 * 2.00 + 0.5 * 4.00 *2.002         =   8.00   rad   =   8.00* 180 /   =   458.600          Finalposition         =   0   +      =   57.3+458.6   =   515.900          Finalposition         =   0   +      =   57.3+458.6   =   515.900

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