Why Is the Angular Velocity of a Clockwise-Spinning CD in the -y Direction?

AI Thread Summary
The discussion centers on determining the direction of angular velocity for a clockwise-spinning CD. The key point is that the answer is given as the -y direction, which some participants find confusing. Using the right-hand rule, the thumb points in the negative y-direction when considering the vertical y-axis as the axis of rotation. This interpretation aligns with the two-dimensional perspective of the problem. Understanding the application of the right-hand rule in this context clarifies why the angular velocity is considered to be in the -y direction.
Qnslaught
Messages
8
Reaction score
0

Homework Statement


A CD player spins a disc in a clockwise direction when looking at the player from the top. What is the direction of the angular velocity.


Homework Equations





The Attempt at a Solution


OK, the key for this practice exam says the answer is in the -y direction. I used the right hand rule to see that the direction is negative, but I thought it would be in the -z direction. Can someone explain why it is -y?
 
Physics news on Phys.org
I believe they have made the answer this way as you are only using two dimensions. If you take the vertical y-axis as the axle for the rotation, your right thumb will be pointing in the negative y-direction when you use the right hand rule.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top