A 48.5-cm diameter disk rotateswith a constant angular acceleration of 2.4 rad/s

Question

A 48.5-cm diameter disk rotateswith a constant angular acceleration of 2.4 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

In general (t)= o+ ot +0.5t2 (t)= t + o o =0 o=57.3 or 1 radian; Now we have (t)= 0.5t2 + 1 (t)= t   where =2.4 rad/s (a) Find the angular speed of the wheelat t = 2.30 s.
(t)= t   (2.30)= 2.4 x 2.30= 5.5rad/s

(b) Find the linear velocity and tangential acceleration of Pat t = 2.30 s.
linear velocity V= /R V= 5.5/(0.485)=2.66 m/s tangential acceleration a=R a=2.4 x 0.485 = 1.16m/s2
c) Find the position of P (in degrees, with respect to the positivex-axis) at t = 2.30s.
(t)= 0.5t2 + 1 (2.30)= 0.5(2.4)(2.30)2 + 1 (2.30)= 7.35 rad = 420 °
(t)= 0.5t2 + 1 (t)= t   where =2.4 rad/s (a) Find the angular speed of the wheelat t = 2.30 s.
(t)= t   (2.30)= 2.4 x 2.30= 5.5rad/s

(b) Find the linear velocity and tangential acceleration of Pat t = 2.30 s.
linear velocity V= /R V= 5.5/(0.485)=2.66 m/s tangential acceleration a=R a=2.4 x 0.485 = 1.16m/s2
c) Find the position of P (in degrees, with respect to the positivex-axis) at t = 2.30s.
(t)= 0.5t2 + 1 (2.30)= 0.5(2.4)(2.30)2 + 1 (2.30)= 7.35 rad = 420 °

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