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

A person stands in contact against the wall of a cylindrical drum of radius R rotating with an angular velocity ω . If the coefficient of friction between the wall and the person is μ , the minimum rotational speed of the cylinder which enables the person to remain stuck to the wall when the floor is suddenly removed is

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

The correct option is A ωmin=gμR
The person will remain stuck to the wall if the limiting static frictional force acting between floor and the person
must balance its own weight.


Normal reaction provides the necessary centripetal force to the person.

N=Mω2R

(fs)max=μN=μMω2R

or, applying condition of vertical equilibrium

(fs)max=Mg

μMω2R=Mg

When friction acting is maximum, that will correspond to minimum angular velocity of drum such that slipping is prevented.

ω2min=gμR

ωmin=gμR

flag
Suggest Corrections
thumbs-up
4
similar_icon
Similar questions
View More
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon