CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A string wrapped around the pulley as shown in the figure is pulled with a constant force F of magnitude 100 N. The radius R and moment of inertia I of the pulley are 0.1 m and 1.25×103 kgm2 respectively. If the string does not slips, what is the angular velocity of pulley after 1 m of string has been unwounded? Assume the pulley starts from rest.



A
400 rad/sec
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
200 rad/sec
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
800 rad/sec
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
100 rad/sec
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 400 rad/sec

From the FBD, force on the bearing of the pulley (B) and weight of the pulley (Mg) does not exert a torque around the rotational axis, therefore no work is done by them on the pulley.
As pulley rotates through an angle θ,F act through a distance d such that d=Rθ.
Since, the torque due to F has magnitude
τ=R×F
We have, work done, W=τθ=(R×F)×(θ)
W=F.d.
If the force on the string acts through a distance 1 m, we have, from the work energy theorem,
W=KE2KE1
F.d=12Iω2212Iω21 [but ω1=0]
(100)(1)=12(1.25×103)ω22
ω2=400 rad/sec
So, the angular velocity of pulley after 1 m of string has been unwounded is 400 rad/sec

flag
Suggest Corrections
thumbs-up
2
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon