Question

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.

Answer 1

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
\[\tau = I \alpha \\ TR = I\frac a R \\ T = I \frac{a}{R^2}\]

Answer 2

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 \(\ddot x = \alpha g\)

Answer 3

is this something like this??

Answer 4

http://www.maplesoft.com/content/EngineeringFundamentals/4/MapleDocument_30/Rotation%20MI%20and%20Torque.pdf

Answer 5

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

\(L = T - U = \frac{1}{2} \dfrac{I}{R^2} \dot x^2 + \frac{1}{2}m \dot x^2 +mgx\)

\(L_{\dot x} = \dfrac{I}{R^2} \dot x +m \dot x, \; L_{x} = mg\)

\((L_{\dot x})^\prime =L_x\)

\(\left (\dfrac{I}{R^2} + m \right)\ddot x = mg\)

Answer 6

\( T = m(g-a)\)
\(T = I \frac{a}{R^2} = \dfrac{1}{2}Ma\)

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

Answer 7

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

\(v = \dfrac{2m}{M+2m}g \; t\)

\(\dot \theta = \dfrac{1}{R}\dfrac{2m}{M+2m}g \; t\)

\(E_D = \frac{1}{2} I \dot \theta ^2 = \frac{1}{2} \left(\frac{1}{2}MR^2 \right) \left[\dfrac{1}{R}\dfrac{2m}{M+2m}g \; t\right]^2 = ....\)