Problem rigid body kinematics c 45 j i vecvc 2ms 1 vecac...

Problem. Rigid Body Kinematics: C 45° j i $\vec{V_c} = 2ms^{-1}$ $\vec{a_c} = 10ms^{-2}$ Given $|\vec{AB}| = 0.6m$ $|\vec{BC}| = 0.4m$ $\vec{V_c} = 2 \vec{i} ms^{-1}$ $\vec{a_c} = 10 (-\vec{i}) ms^{-2}$ To find $\omega_{BC} = ?$ $\alpha_{BC} = ?$ B $\omega_{AD}$ 60° A

Transcript

00:01 So for problem 15 .1 .86, we'll be given two components of the angular velocity of point a of the plate.

00:09 We'll be given components x and y.

00:12 And we'll be given the x component of the angular velocity.

00:16 And we've been asked to calculate the angular velocity and the linear velocity of point d of the plate.

00:24 The strategy here is basically to find a way to relay the angular velocity with the linear velocities for points a and d and then solve for the angular velocity.

00:38 Two comments before we begin.

00:41 Firstly, the drawing might be a little bit deceiving.

00:45 We shouldn't infer from the drawing that the angular velocity points towards the y direction.

00:56 And also, it must be clear from the drawing and the description that there's no translational motion here.

01:05 So actually the relation that relates the linear velocity to the angular velocity is quite straightforward.

01:17 Typical relation, cross -product, the angular velocity to the position vector of the point.

01:27 Here this is for point a and correspondingly for point d.

01:31 We'll do this later.

01:37 There's no here there's no linear velocity for me for the for the origin since there's no translational motion.

01:53 So now from the join, someone can see that the position vector of point a with respect to the origin is this one.

02:18 And this is in millimeters.

02:28 And therefore we can express the cross product in the form of determinant for convenience.

02:39 This is again the linear velocity of point a, which equals to the crossbarate of the angular velocity times the position vector...

Question

Problem. Rigid Body Kinematics: C 45° j i $\vec{V_c} = 2ms^{-1}$ $\vec{a_c} = 10ms^{-2}$ Given $|\vec{AB}| = 0.6m$ $|\vec{BC}| = 0.4m$ $\vec{V_c} = 2 \vec{i} ms^{-1}$ $\vec{a_c} = 10 (-\vec{i}) ms^{-2}$ To find $\omega_{BC} = ?$ $\alpha_{BC} = ?$ B $\omega_{AD}$ 60° A

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