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ganeshie8

  • one year ago

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.

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  1. ganeshie8
    • one year ago
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    |dw:1443699764703:dw|

  2. BAdhi
    • one year ago
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    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 \[\tau = I \alpha \\ TR = I\frac a R \\ T = I \frac{a}{R^2}\]

  3. IrishBoy123
    • one year ago
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    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 \(\ddot x = \alpha g\)

  4. anonymous
    • one year ago
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  5. anonymous
    • one year ago
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    is this something like this??

  6. IrishBoy123
    • one year ago
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    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 \(L = T - U = \frac{1}{2} \dfrac{I}{R^2} \dot x^2 + \frac{1}{2}m \dot x^2 +mgx\) \(L_{\dot x} = \dfrac{I}{R^2} \dot x +m \dot x, \; L_{x} = mg\) \((L_{\dot x})^\prime =L_x\) \(\left (\dfrac{I}{R^2} + m \right)\ddot x = mg\)

  7. ganeshie8
    • one year ago
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    \( T = m(g-a)\) \(T = I \frac{a}{R^2} = \dfrac{1}{2}Ma\) 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

  8. IrishBoy123
    • one year ago
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    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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