I spoke with my teacher and he said there was a typo in the answer selection. The correct answer is in fact 1.36 m/s and answer B should be 1.36 m/s. Thank you dauto and NascentOxygen.
As a side note: he also mentioned that the acceleration he gave was the tangential acceleration. Luckily the...
Ok, I set up my equation: ∑Fx=fs=mα→ μmg=m\sqrt{ar2+at2} (I'm not sure why itex isn't working for me, so i omitted the commands). After some simplification I get ω2=\sqrt{(μg)2-(αr)2}/r.
I plugged in my values at this point and found that:
ω2≈3.264rad/s → ω≈1.806rad/s.
Then i plugged in...
I'm a bit confused. Are you saying that ∑Fx=fs=mar should actually be ∑Fx=fs=mα? then I take the squareroot ar2+at2 plug it into my force equation and then solve for ω?
Wow, thank you. I didn't even notice my units were different than the given answers. I converted my units to m/s using ##v=\omega r##. This gave me 1.356m/s which is between answers B and C. This should be the speed of the object just before it starts to slip.
Did I approach this problem...
Homework Statement
A 75g mass sits 75cm from the center of a rotating platform undergoing a uniform angular acceleration of 0.125rad/s . The coefficient of static friction between the mass and the platform is 0.250. What is the speed of the mass when is slides off?
A. 0.889 m/s
B. 1.26...