A 41.7-cm diameter disk rotates with aconstant angular acceleration of 2.5rad/s

Question

A 41.7-cm diameter disk rotates with aconstant angular acceleration of 2.5rad/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 positivex-axis) at t = 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 positivex-axis) at t = 2.30s.
4° linear velocity 2 m/s tangential acceleration 3 m/s2

Explanation / Answer

diameter = 41.7 cm, radius r = 20.85 cm = 0.2085 m, constant angular acceleration = 2.50 rad /s2, initial angular velocity = 0, initial position of P = o = 57.3o= 1 rad At t = 2.30 s, A) the angular speed of the wheel = t = 5.75rad/s B) the linear velocity of P: v = r = 1.198875 m/s tangential acceleration of P: at = r = 0.52125m/s2 C) = o + t2/2 = 6.6125 rad= 379.06o(It appears as 379.06 - 360 =19.06o)

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