Practice portal › Rotational Motion › Rotational Dynamics
Asked in JEE Main 9th April 1st Shift 2024 · Pulleys, strings and hanging masses
The string pulls tangentially at the rim, so τ=FR=(40)(0.1)=4 N m.
α=τ/I=4/(0.40)=10 rad s⁻².
The force is steady, so α is constant and, starting from rest, ω=α t.
ω=(10)(10)=100 rad s⁻¹.
x=100.
Reading this solution needs no sign-in. You only sign in to practise against the clock and keep your progress.
Practise this with a timer