It is. However, tension T was not one of the variables we were allowed to express in our answer. That's why Tsintheta/m won't work here. Thank you for the inquiry.
@@driftingthroughthisplace5898 college board just likes making their wording inconsistent and overly complex.. if you look at the answer key, they don't have m in the solution either so it isn't necessary but you still can to be safe
This very well explained, considering I understood it. Thank you for making these videos, you're a life saver!
thank you so much for all of these videos, i was out for a week sick and they are really helping me catch up
When AP classroom died, this man has saved my life
You’re my new teacher 😂
Isn’t the centripetal acceleration equal to Tsintheta/m?
It is. However, tension T was not one of the variables we were allowed to express in our answer. That's why Tsintheta/m won't work here. Thank you for the inquiry.
Very interesting
Good explanations to questions. 👍
How does the dividing mac=Tsin(theta by mg=Tcos(theta work? Is it because they both have the shared variable of T?
mac = Tsintheta, mg = Tcostheta, therefore T = mg/cos(theta), therefore Tsintheta = mg/cos(theta) * sin(theta) = mgtan(theta)
basically since mg = Tcostheta, its like dividing with the same number on both sides
Why there isn't a normal force acting on the block?
Normal force only applies when an object is on a surface
Your derivation didnt include M, and the question says in terms of M theta and physical constants
M cancels out.
@@pushkara1483 I'm aware, but the problem still asks for it in terms of M. I didn't cancel the Ms out and still got it correct
@@driftingthroughthisplace5898 yea you're right but you don't need mass, you'll still get it right and it makes more sense to cancel out mass
@@driftingthroughthisplace5898 college board just likes making their wording inconsistent and overly complex.. if you look at the answer key, they don't have m in the solution either so it isn't necessary but you still can to be safe