Point A of the circular disk is at the angular position theta = 0 at time t = 0.

Question

Point A of the circular disk is at the angular position theta = 0 at time t = 0. The disk has angular velocity omega0 = 0.1 rad/s at t = 0 and subsequently experiences a constant angular acceleration alpha = 2 rad/s2. Determine the velocity and acceleration of point A interms of fixed i and j unit vectors at time t = 1 s. Repeat Prob. 5/20 except now the angular acceleration of the disk is given by alpha = 2t, where t is in seconds and alpha is in radians per second squared. Determine the velocity and acceleration of point A in terms of fixed i and j unit vectors at time t = 2s. Repeat prob 5/20, except now the angular acceleration of the disk is given

Explanation / Answer

ang.acc alpha = d w /dt

2t = dw/dt

Integrating both the sids with respect to t,

t^2 = w + C1

At t = 0, we have w = w0 = 0.1

Hence, C1 = -0.1

Thus t^2 = w - 0.1.................(1)

Now, w = d theta / dt

t^2 = d theta / dt - 0.1

Integrating both the sides with respect to t,

t^3/3 = theta - 0.1*t + C2

At t = 0, we have theta = 0.

Hence, C2 = 0

Thus, t^3 /3 = theta - 0.1*t...............(2)

From eqn 1, At t = 2 s, angular velocity w = 2^2 + 0.1 = 4.1 rad/s

From eqn 2, At t = 2 s, angle theta = 2^3 /3 + 0.1*2 = 2.86667 rad = 164.25 deg

Direction of tangential vel and acceleration = theta + 90 = 164.25 + 90 deg = 254.25 deg

At t = 2 s, linea velocity = rw = 0.2*4.1 = 0.82 m/s

At t = 2 s, tangential acceleration a_t = r*alpha = 0.2*(2*2) = 0.8 m/s2

At t = 2 s, normal acceleration a_n = rw^2 = 0.2*4.1^2 = 3.362 m/s2

In i-j vectors form,

v = 0.82 Cos 254.25 i + 0.82 Sin 254.25 j = -0.226 i - 0.789 j

Tangential acceleration a_t = 0.8 Cos 254.25 i + 0.8 Sin 254.25 j = -0.217 i - 0.77 j

Normal acceleration a_n = 3.362 Cos 164.25 i + 3.362 Sin 164.25 j = -3.236 i + 0.9126 j

Net acceleration = a_t + a_n = -3.453 i + 0.1426 j

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