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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.
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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ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2dw:1443699764703:dw

BAdhi
 one year ago
Best ResponseYou've already chosen the best response.2to 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> \[ \downarrow F = ma \\ mg  T = ma \\ T = m(ga)\] to M \[\tau = I \alpha \\ TR = I\frac a R \\ T = I \frac{a}{R^2}\]

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2or 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 \(\ddot x = \alpha g\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0is this something like this??

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2or 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
 one year ago
Best ResponseYou've already chosen the best response.2\( T = m(ga)\) \(T = I \frac{a}{R^2} = \dfrac{1}{2}Ma\) 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
 one year ago
Best ResponseYou've already chosen the best response.2no 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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