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

A massless cord is wound round the circumference of the wheel of radius r. The axis of the wheel is horizontal and the moment of inertia about the centre is I. A weight mg is attached to the cord at the end. The weight falls from rest and performs pure rolling. After falling through distance h, angular velocity of wheel is

A
2ghr
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
mghI+2mr2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
2mghI+mr2
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
2mghI+2mr2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C 2mghI+mr2
Loss of potential energy= gain in kinetic energy
Loss of potential energy=mgh
Gain of kinetic energy = translational kinetic energy + rotational kinetic energy
Gain in Kinetic energy = 12mv2+12Iω2
mgh=12mv2+12Iω2
on solving above equation, we get
ω2[mr2+I2]=mgh
[2mghmr2+I]1/2=ω

flag
Suggest Corrections
thumbs-up
1
similar_icon
Similar questions
View More
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Relative
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon