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A disk of mass 4kg is mounted on a massless rigid rod BC as shown in the diagram. A torque $τ = 5 e^{0.5 t}$ is applied to the arm at C.

Consider three cases:
Case-I: The disk is free to rotate about its bearing at B.
Case-II: The disk is rigidly attached to its bearing at B.
Case-III: The disk is freely rotating about its bearing at B with angular velocity $(- 10 \hat{k})$ rad/s prior to application of torque. Take : e = 2.7

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In case-I, the kinetic energy of disk at t = 2 sec is $\frac{289}{2} J .$

In case-II, the kinetic energy of disk at t = 2 sec is $\frac{1156}{9} J .$

In case-III, the kinetic energy of disk at t = 2 sec is $\frac{6281}{16} J .$

In case-III, the kinetic energy of disk at t = 2 sec is $\frac{603}{4} J .$

answer is A, B, C.

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Detailed Solution

$Torque about z-axis = 5 e^{0.5 t}$

$I_{z} α = 5 e^{0.5 t}$

$α = \frac{5}{I_{z}} e^{0.5 t}$

$\frac{d ω}{dt} = \frac{5}{I_{z}} e^{0.5 t}$

${(ω_{f} - ω_{i})}_{about z} = \frac{10}{I_{z}} (e^{0.5 t} - 1)$

$At t = 2 \sec$

${(ω_{f} - ω_{i})}_{about z} = \frac{10}{I_{z}} (e - 1)$

${(ω_{f} - ω_{i})}_{about z} = \frac{17}{I_{z}}$

Case (I)

$I_{z} = m_{disc} (0.5)^{2} = 1 {kgm}^{2}$

$ω_{z} = 17 rad / s^{2}$

$KE = \frac{1}{2} I ω^{2} = \frac{1}{2} \times 1 \times (17)^{2} = \frac{289}{2} J$

Case (II)

$I_{z} = m_{disc} \frac{R^{2}}{2} + m_{disc} (0.5)^{2}$

$I_{z} = \frac{9}{8} {kgm}^{2}$

$ω_{z} = \frac{136}{9} rad / s$

$KE = \frac{1}{2} \times \frac{9}{8} \times {(\frac{136}{9})}^{2} = \frac{1156}{9} J$

Case (III)

$ω_{z} = 17 rad / s$

$ω about axis of the disc = 10 rad / s$

$\begin{array}{l} KE = \frac{1}{2} I_{disc} ω_{disk}^{2} + \frac{1}{2} I_{z} ω_{z}^{2} \\ KE = (\frac{1}{2} \times \frac{1}{8} \times (10)^{2}) + (\frac{1}{2} \times 1 \times (17)^{2}) \\ KE = \frac{603}{4} J \end{array}$

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