Search any question & find its solution
Question:
Answered & Verified by Expert
A constant torque of $100 \mathrm{~N} \mathrm{~m}$ turns a wheel of moment of inertia $300 \mathrm{~kg} \mathrm{~m}^2$ about an axis passing through its centre. Starting from rest, its angular velocity after $3 \mathrm{~s}$ is
Options:
Solution:
1755 Upvotes
Verified Answer
The correct answer is:
$1 \mathrm{rad} / \mathrm{s}$
About fixed axis $\rightarrow$ Torque $=I \alpha$
$\begin{aligned} & \alpha=\frac{100}{300} \\ & \alpha=\frac{1}{3} \mathrm{rad} / \mathrm{s}^2\end{aligned}$
$\begin{aligned} & \Rightarrow \omega_i=0 \\ & \omega_f=\omega_i+\alpha t \\ & \text { at } t=3 \mathrm{~s} \\ & \omega_f=\frac{1}{3} \times 3=1 \mathrm{rad} / \mathrm{s}\end{aligned}$
$\begin{aligned} & \alpha=\frac{100}{300} \\ & \alpha=\frac{1}{3} \mathrm{rad} / \mathrm{s}^2\end{aligned}$
$\begin{aligned} & \Rightarrow \omega_i=0 \\ & \omega_f=\omega_i+\alpha t \\ & \text { at } t=3 \mathrm{~s} \\ & \omega_f=\frac{1}{3} \times 3=1 \mathrm{rad} / \mathrm{s}\end{aligned}$
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.