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.
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.
°
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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