Physics · SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
हिन्दीIn the system shown below, a light cord is wrapped around the rim of a uniform flywheel of mass \( M = 20 \text{ kg} \) and radius \( R = 20 \text{ cm} \) (mounted on a frictionless horizontal axle). A steady force \( F = 25 \text{ N} \) pulls the cord vertically downwards. If the cord unwinds for a duration such that the wheel makes \( 5 \) complete revolutions from rest, the work done by the pull is:
View Answer & Solution
Correct answer · Option A
\( 50\pi \text{ J} \)
Explanation
Length of cord unwound in 5 revolutions \( s = 5 \times (2\pi R) = 5 \times 2\pi(0.20) = 2\pi \text{ m} \). Work done \( W = F \times s = 25 \times 2\pi = 50\pi \text{ J} \).