Instantaneous center of 0 velocity

[sami23]

Homework Statement

At the instant shown, determine omega_BC, the magnitude of the angular velocity of link BC. Use the values theta = 32.0 degrees v_C = 4.00 ft/s, L_1 = 3.60 ft, and L_2 = 0.700 ft.

Homework Equations

v_B = v_A + omega x r_BA

The Attempt at a Solution

I found IC and tried to calculate length r_c/ic and r_b/ic using the sine law:
L_2 / sin(32) = r_b/ic / sin (90-32)
r_b/ic = (0.7 * sin(58) ) / sin(32) = 1.1202 ft

L_2 / sin(32) = r_c/ic / sin(90)
r_c/ic = 0.7 *1 / sin(32) = 1.32096 ft

using the formula omega_bc = v_c / r_c/ic:
omega_bc = 4/1.32096 = 3.028 rad/s but it's wrong.

Please help.

 

[mark726]

Thank you for your question. In order to determine the magnitude of the angular velocity of link BC, we need to first calculate the linear velocity of point C using the given information. We can use the equation v_C = v_B + omega x r_BC, where r_BC is the distance between points B and C. Since we know the value of v_C and the angle between v_B and r_BC, we can use trigonometry to calculate the magnitude of v_B.

Using the law of cosines, we can find the length of r_BC:
r_BC^2 = L_1^2 + L_2^2 - 2(L_1)(L_2)cos(90-32)
r_BC = 3.66 ft

Now, using the law of sines, we can find the magnitude of v_B:
v_B / sin(32) = 4 / sin(90)
v_B = 4 * sin(32) = 2.08 ft/s

Finally, we can use the equation omega_bc = v_c / r_c/ic to find the magnitude of the angular velocity:
omega_bc = 4 / (3.66 * sin(32)) = 0.355 rad/s

I hope this helps. Let me know if you have any further questions or if you need clarification on any of the steps. Keep up the good work!