ganeshie8
  • ganeshie8
Find the rotational kinetic energy of the disk shown in the figure \(2.5\) seconds after it starts accelerating. Given : mass of disk, M = 2.5kg radius of disk = 20cm mass of block, m = 1.2kg Assume there is no friction between rope and the disk. Also, assume the rope is massless.
Calculus1
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SOLVED
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jamiebookeater
  • jamiebookeater
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ganeshie8
  • ganeshie8
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BAdhi
  • BAdhi
to find the rotational kinetic energy you need to find the \(\omega\) at \(t =2.5s\) for that you need the acceleration. m will have a constant acceleration of a and the M also will have the same acceleration on the circumference. to m-> \[ \downarrow F = ma \\ mg - T = ma \\ T = m(g-a)\] to M \[\tau = I \alpha \\ TR = I\frac a R \\ T = I \frac{a}{R^2}\]
IrishBoy123
  • IrishBoy123
or use conservation of Energy \(E = \frac{1}{2} I \dot \theta^2 + \frac{1}{2}m \dot x^2 - mgx\) \(\dot E = 0 = I \dot \theta \ddot \theta+ m \dot x \ddot x - mg \dot x\) \(x = R \theta, \dot x = R \dot \theta \) etc \( \dfrac{I}{R^2} \dot x \ddot x+ m \dot x \ddot x - mg \dot x = 0\) \(\left (\dfrac{I}{R^2} + m \right)\ddot x - mg = 0\) same answer \(\ddot x = \alpha g\)

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anonymous
  • anonymous
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anonymous
  • anonymous
is this something like this??
anonymous
  • anonymous
http://www.maplesoft.com/content/EngineeringFundamentals/4/MapleDocument_30/Rotation%20MI%20and%20Torque.pdf
IrishBoy123
  • IrishBoy123
or Lagrange \(L = T - U = (\frac{1}{2} I \dot \theta^2 + \frac{1}{2}m \dot x^2) - (-mgx)\) but \(\theta \) and x are dependent so this becomes an expression in x \(L = T - U = \frac{1}{2} \dfrac{I}{R^2} \dot x^2 + \frac{1}{2}m \dot x^2 +mgx\) \(L_{\dot x} = \dfrac{I}{R^2} \dot x +m \dot x, \; L_{x} = mg\) \((L_{\dot x})^\prime =L_x\) \(\left (\dfrac{I}{R^2} + m \right)\ddot x = mg\)
ganeshie8
  • ganeshie8
\( T = m(g-a)\) \(T = I \frac{a}{R^2} = \dfrac{1}{2}Ma\) together imply \(a =\dfrac{2mg}{2m+M} \) so, \(\omega(2.5) = 2.5*\alpha = 2.5*\dfrac{a}{R}\) Nice @BAdhi looks conversation of energy also requires me to find \(T\) and \(a\) first hmm @IrishBoy123
IrishBoy123
  • IrishBoy123
no need for T in terms of acceleration, sort of.... \[\ \ddot x = \dfrac{2m}{M+2m}g = const.\] so use the equations of motion for constant acceleration, eg \(v = u + a t \) to find velocity and angular velocity at t = 2.5 \(v = \dfrac{2m}{M+2m}g \; t\) \(\dot \theta = \dfrac{1}{R}\dfrac{2m}{M+2m}g \; t\) \(E_D = \frac{1}{2} I \dot \theta ^2 = \frac{1}{2} \left(\frac{1}{2}MR^2 \right) \left[\dfrac{1}{R}\dfrac{2m}{M+2m}g \; t\right]^2 = ....\)

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