Search any question & find its solution
Question:
Answered & Verified by Expert
A disc of radius $0.4 \mathrm{~m}$ and mass $1 \mathrm{~kg}$ rotates about an axis passing through its center and perpendicular to its plane. The angular acceleration of the disc is $10 \mathrm{rad} / \mathrm{s}^2$. The tangential force applied to the rim of the disc is
Options:
Solution:
1292 Upvotes
Verified Answer
The correct answer is:
$2 \mathrm{~N}$
$\mathrm{R}=0.4 \mathrm{~m}, \mathrm{M}=1 \mathrm{~kg}, \alpha=10 \mathrm{rad} / \mathrm{s}^2$
Moment of inertia, $\mathrm{I}=\frac{\mathrm{MR}^2}{2}=\frac{1 \times(0.4)^2}{2}=0.08 \mathrm{~kg} \mathrm{~m}^2$
Torque, $\tau=\mathrm{RF}=\mathrm{I} \alpha$
$\therefore \mathrm{F}=\frac{\mathrm{I} \alpha}{\mathrm{R}}=\frac{0.08 \times 10}{0.4}=2 \mathrm{~N}$
Moment of inertia, $\mathrm{I}=\frac{\mathrm{MR}^2}{2}=\frac{1 \times(0.4)^2}{2}=0.08 \mathrm{~kg} \mathrm{~m}^2$
Torque, $\tau=\mathrm{RF}=\mathrm{I} \alpha$
$\therefore \mathrm{F}=\frac{\mathrm{I} \alpha}{\mathrm{R}}=\frac{0.08 \times 10}{0.4}=2 \mathrm{~N}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.