A 47.4-cm diameter disk rotates with a constant angular acceleration of 2.80 rad

Question

A 47.4-cm diameter disk rotates with a constant angular acceleration of 2.80 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.

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

Explanation / Answer

here,

diameter , d = 47.4 cm

radius , r = 0.237 m

accelration , a = 2.8 rad/s^2

theta = 57.3 degree

(a)

at t = 2.3 s

the angular speed , w = w0 + a*t

w = 0 + 2.8 * 2.3

w = 6.44 rad/s

the angular speed at t = 2.3 s is 6.44 rad/s

(b)

the linera velocity , v = r*w

v = 1.53 m/s

the tangential accelration , a' = r * a

a' = 0.66 m/s^2

(c)

theta' = w0*t + 0.5 * a*t^2

theta' = 0 + 0.5 * 2.8 * 2.3^2

theta' = 7.41 rad

theta' = 424.55 degree

angle of P = theta' + theta

P = 481.85 degree

the position of point P is 121.85 degree from the positive x axis

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