6. -18 points SerCP11 7 P012 My Notes Ask Your Teacher A 40.0-cm diameter disk r

Question

6. -18 points SerCP11 7 P012 My Notes Ask Your Teacher A 40.0-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) At t 2.35 s, find the angular speed of the wheel. rad/s (b) At t 2.35 s, find the magnitude of the linear velocity and tangential acceleration of P. linear velocity tangential acceleration m/s m/s2 (c) At t-2.35 s, find the position of P (in degrees, with respect to the positivex-axis) o counterclockwise from the +x-axis Need Help? Read

Explanation / Answer

Solution:

a) = angular acceleration x time

=> = 2.80 x 2.35

=> = 6.58 rad /s

b) Using V = * r

=> V = 6.58 x 0.20

=> V = 1.316 m/s

and , a = angular acc. * r

=> a = 2.80 * 0.20

=> a = 0.56 m/s^2

So, linear velocity = 1.316 m/s

and tangential acceleration = 0.56 m/s^2

c) P = 57.3 x 3.14 / 180 + 0.5 * 2.80 * (2.35)^2

=> P = 8.731 rad

=> P = 140.5 degree

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