A 51.3-cm diameter disk rotates with a constant angular acceleration of 2.4 rad/

Question

A 51.3-cm diameter disk rotates with a constant angular acceleration of 2.4 rad/s2. It starts from rest at t = 0, and a line drawn from the center of the disk to a point P on the rim of the disk makes an angle of 57.3° with the positive x-axis at 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 P at t = 2.30 s.
linear velocity m/s
tangential acceleration m/s2

c) Find the position of P (in degrees, with respect to the positive x-axis) at t = 2.30s.
°

Explanation / Answer

GIVEN: The diameter of the disk is r = 51.3  cm = 51.3 x 10-2m The angular acceleration of the disk is = 2.4 rad/s2 The line drawn from the center of the disk to a point P on the rim of the disk makes an angle = 57.3o with the positive x-axis at this time. We know that 1 radian = 57.30o or = 1 radian ---------------------------------------------------------------------------  a)The angular speed of the wheel at t = 2.30s is w = wo + t Here,wo= 0 rad/s or w = 0 +(  2.4 rad/s2 ) (  2.30 s ) = 5.52  rad/s -------------------------------------------------------------------------------  b)The linear velocity at of P at t = 2.30s is v = ( r ) ( w )  v = ( 51.3 x  10-2  m )  ( 5.52  rad/s )  = 2.83  m/s The tangential acceleration of P at t = 2.30s is T= r * or T= ( 51.3 x  10-2  m )  (   2.4 rad/s2  )  = 1.23rad/s2 -----------------------------------------------------------------------------------  c)The position of P at t = 2.30s is = wot + (1/2)t2 or = 0 + (1/2) (   2.4 rad/s2  )  (2.30 s )2 = 6.348 radian But 1 radian = 57.30o So from initial position ( 1 rad ) it makes 7.348 radians or = 7.348  x  57.30o = 421.04 deg or  421.04 -360 = 60.04 deg   with  x-axis                                                                     or   approx  60 deg             

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