Magnitude and direction of velocity

prabhjyot
Messages
10
Reaction score
0
The end A of the link in a guide has a constant downward velocity of 3 m/s as shown in the diagram. For the instant where the link AB makes an angle of 75 degrees with the vertical guide calculate the magnitude and direction of the velocity of the roller at B, the magnitude and direction of velocity of the centre of the link AB and the angular velocity of the link AB. Is the velocity at B also constant?
 

Attachments

  • dia.gif
    dia.gif
    1.7 KB · Views: 554
Physics news on Phys.org
That looks like a typical Calculus "related rates" problem!

You need two formulas: by the Pythagorean theorem, a2+ b2= 2002, where b is the distance from B to the corner and a is the distance from A to the corner. Differentiate that formula with respect to time (using the chain rule, of course) to determine db/dt (you are given da/dt), the "magnitude of the velocity of the roller at B". The direction should be obvious.

You also have cos(\theta)= \frac{a}{200} where \theta is the angle AB makes with the vertical guide (75 degrees at the instant in question). Differentiate that to determine "the angular velocity of the line AB", \frac{d\theta}{dt}.

For the midpoint, you can do the same thing but use 100 mm instead of 200.
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top