Physics Help Forum Angular motion problem

 Periodic and Circular Motion Periodic and Circular Motion Physics Help Forum

 Oct 13th 2017, 05:37 PM #1 Junior Member   Join Date: Oct 2017 Posts: 8 Angular motion problem Problem A long string is wrapped around a 5.00-cm-diameter cylinder, initially at rest, that is free to rotate on an axle. The string is then pulled with a constant acceleration of 1.50 m/s2 until 1.25 m of string has been unwound. Assume the string unwinds without slipping. (a) What is the angular displacement of the cylinder after 1.25 m of string has been unwound? (b) What is the angular acceleration of the cylinder as the string is unwound? (c) What is the final angular velocity of the cylinder, expressed in rpm? Knowns tangential acceleration(at) =1.25m/s arc length(s)=1.25m radius of the cylinder(r) = .025m Attempt at solving the problem a.) Θ=s/r Θ=1.25/.025 Θ=50 radians b.) at=αr 1.25=α(.025) 1.25/.025=α a=60 radians/sec^2 c.) Using the equation wf=wi+2αΘ wf=2αΘ wf=2(60)(50) wf=6000rads/second This is where I am stuck. I'm not really quite sure if my reasoning that the tangential acceleration is 1.25m/s^2, or if my final angular velocity is reasonable at all. I'm also having trouble wrapping my head around the conversion to RPM. Thanks for any help in advance.
 Oct 13th 2017, 06:43 PM #2 Senior Member     Join Date: Aug 2008 Posts: 113 (a) ok (b) $a = 1.50 \, m/s^2$ you typed in 1.25, but arrived at the correct solution (c) $\omega_f^2 = \omega_0^2 + 2\alpha \Delta \theta$ conversion of radians/sec to revolutions/min ... $\dfrac{rad}{sec} \cdot \dfrac{1\, rev}{2\pi \, rad} \cdot \dfrac{60 \, sec}{1 \,min} = \dfrac{rev}{min}$
 Oct 13th 2017, 08:16 PM #3 Junior Member   Join Date: Oct 2017 Posts: 8 Wow I can't believe that I forgot that angular velocity was squared, thanks!

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