Physics Help Forum Energy in rotational motion

 Kinematics and Dynamics Kinematics and Dynamics Physics Help Forum

 May 17th 2017, 04:06 AM #1 Member   Join Date: Jan 2015 Posts: 96 Energy in rotational motion Hi everyone, I have come across a problem which I initially thought was straightforward but now I find that no matter how I try to solve it I cannot get the correct answer that is shown at the back of my Physics textbook. I just can't see where I am going wrong. PROBLEM A friction less pulley has a shape of a uniform solid disk of mass 2.50 kg and a radius of 20.0 cm. A 1.50 kg stone is attached to a very light wire that is wrapped around the rim of the pulley, and the system is released from rest. a) How far must the stone fall so that the pulley has 4.50 J of Kinetic Energy ? there is a part b) to this problem but I will omit it as it is part a) that I have the problem with. Initially I found the Moment of Inertia of the pulley using .. I = 1/2 MR(sqrd), where I is it's Moment of Inertia and M is the mass of the pulley and R is the radius of the disk (I took the disk to be a solid cylinder). from this I = 0.05 kgm(sqrd) I then used this value for Moment of inertia to find rotational velocity (w), using .. K = I w(sqrd) where K is the Kinetic energy of the pulley (=4.50 J) and w = rotaional velocity. from this w = 13.4 rad/s I then found the a value for the rotational acceleration .. a(tan) - r. alpha where a(tan) is the tangential acceleration of the pulley and alpha the rotational acceleration of the pulley and noting that a(tan) is equal to the acceleration of the stone under gravity so a(tan) = 9.8 m's(sqrd) this gives alpha = 49.0 rad/s(sqrd) Next I used w(sqrd) = w(0)(sqrd + 2 alpha (theta-theta(0)) to find theta (the angle in rads that is completed by the rotating pulley) and where w(0) is initial angular velocity (= 0, from rest) and theta(0) initial angle value (= 0 )# from this theta = 18.36 rad finally I used the equation d = r . theta where d = distance traveled, r = radius of pulley and theta is the angle of rotation as before. so I get d= 3.66 m ( 366 cm) However the answer in the book says d = 67.3 cm ? Can anyone point out where I have gone wrong KI can't see it ? Thanks. Jackthehat
 May 17th 2017, 06:55 AM #2 Junior Member   Join Date: May 2017 Posts: 12 not an expert I think i would approach this as GR - Newton's 2nd law using the diameter of the pulley for distance in the acceleration. My first thought was that i would then have to solve for joules of the pulley's acceleration, but if both motions were graphed a rough answer could be found. Since all bodies fall at the same speed much of GR is constant 32ft/(s)(s) and both accelerations are the same (per the wire) the acceleration in Newton's 2nd could be solved this way. that should make the equation simpler. It seems like you will be solving a square root at first glance. a = F / m 32ft/(s)(s) = F / m (or the diameter of the pulley)(125.6cm)
 May 17th 2017, 07:46 AM #3 Senior Member     Join Date: Aug 2008 Posts: 113 initial gravitational potential energy of stone = rotational KE of the pulley + translational KE of the stone $mgh = KE_R + \dfrac{1}{2}mv^2$ $h = \dfrac{KE_R}{mg} + \dfrac{v^2}{2g}$ starting from rest, note $v^2 = 2ah$ $h = \dfrac{KE_R}{mg} + \dfrac{ah}{g}$ $h = \dfrac{KE_R}{m(g-a)}$ to get $a$ ... $\tau_{net} = I \alpha$ $T r = {1}{2}M_p r^2 \cdot \dfrac{a}{r}$ $\color{red}{T = \dfrac{1}{2}M_p a}$ $F_{net} = ma$ $\color{red}{mg - T = ma}$ solving the system ... $a = \dfrac{2mg}{M_p + 2m}$ returning to the energy-based equation for $h$ ... $h = \dfrac{KE_R}{m\left(g-\frac{2mg}{M_p+2m} \right)}$ substituting in the given values for $KE_R$, $M_p$, $m$, and $g=9.8 \, m/s^2$ yields $h=67.3 \, cm$
 May 17th 2017, 09:23 AM #4 Member   Join Date: Jan 2015 Posts: 96 Hi Skeeter, Thanks for taking the time to help me understand the problem a little better. I understand where I went wrong and how to fix it now. Much appreciated. Jackthehat
 May 17th 2017, 09:24 AM #5 Member   Join Date: Jan 2015 Posts: 96 Thanks a lot Matlock for taking the time to help me out .. I think I can give it another go now. Much appreciated. Jackthehat

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