JEE Mains · Physics · STD 11 - System of particles and rotational motion
- JEE Mains 2024
- Hard
- Numerical
A solid circular disc of mass \(50 \mathrm{~kg}\) rolls along a horizontal floor so that its center of mass has a speed of \(0.4 \mathrm{~m} / \mathrm{s}\). The absolute value of work done on the disc to stop it is _______ \(\mathrm{J}\).
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(6\)
Step-by-step Solution
Detailed explanation
Using work energy theorem
\(\mathrm{W}=\Delta \mathrm{KE}=0-\left(\frac{1}{2} \mathrm{mv}^2+\frac{1}{2} \mathrm{I} \omega^2\right)\)
\(\mathrm{W}=0-\frac{1}{2} \mathrm{mv}^2\left(1+\frac{\mathrm{K}^2}{\mathrm{R}^2}\right) \)
\(=-\frac{1}{2} \times 50 \times 0.4^2\left(1+\frac{1}{2}\right)=-6 \mathrm{~J}\)
Absolute work \(=+6 \mathrm{~J}\)
\(W=-6 J \quad|W|=6 J\)
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