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

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

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.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

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)

$ω_{z} = 17 rad / s$

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

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

Watch 3-min video & get full concept clarity

tricks from toppers of Infinity Learn

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

🔥 Start Your JEE/NEET Prep at Just ₹1999 / month - Limited Offer! Check Now!

Get Expert Academic Guidance – Connect with a Counselor Today!

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET