wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A horizontal disc rotating freely about a vertical axis through its centre makes 90 revolutions per minute. A small piece of wax of mass m falls vertically on the disc and sticks to it at a distance r from the axis. If the number of revolutions per minute reduces to 60, the moment of inertia of the disc is

A
mr2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
32mr2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
2mr2
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
3mr2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C 2mr2
Given,
Initial angular velocity of the disc (ω1)=90 rpm
ω1=90×2π60=3π rad/s (revolutions per second)
Initial moment of inertia I1=I
Here, I= MOI of the disc


After piece of wax sticks to the stick:


Final angular velocity of the dic (ω2)
=60 rpm
=60×2π60=2π rad/s
Moment of inertia of the system (disc + wax)
I2=(I+mr2)
As we see, there is no external torque on the system. Hence angular momentum of the system will be conserved.
Li=Lf
I1ω1=I2ω2
I(3π)=(I+mr2)(2π)
i.e I=2mr2

flag
Suggest Corrections
thumbs-up
0
similar_icon
Similar questions
View More
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon