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

A hollow sphere of radius r rolls without slipping in a hemisphere of radius 4r. Calculate the frequency of small oscillations, about point P as shown in figure.


A

No worries! We‘ve got your back. Try BYJU‘S free classes today!
B

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C

No worries! We‘ve got your back. Try BYJU‘S free classes today!
D

None of these

No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B


The hollow sphere is purely rolling on the curvature. Let's assume the sphere is at some θ from the vertical. At this instant its C.O.M has a velocity V. It will be rolling with angular velocity ω such that ω=vr

The centre of the sphere is doing circular motion about the centre of the curvature 0. Centre of sphere is moving with velocity v at distance 3r so its angular velocity about 0 is ω=v3r - - - - - - (1)

Let's write the energy of the sphere at this instant. It will have rotational + translational energy + potential energy.

12mv2 + 12lω2 + mg 3r(1cos θ = E

Translational Rotational Potential

Energy Energy Energy

Now energy = constant

dEdt=0

12m 2vdvdt+d(1223mr2ω2)dtmg 3R sinθdθdt=0 (negative sign as dθdt is decreasing)

( r2ω2=v2) (dθdt=ωω=v3r)

12m 2vdvdt+1223m.2vdvdtmg 3r sin θdθdt=0

mva+23mvamg 3r sin θv3r

53mvamg sin θv=0

53ma=mg sin θ

a=3g sinθ5

If θ was small then angular acceleration of center of mass with respect to center of curvature o will be

α=3g sinθ5(13r)

α=gθ5r

ω=g5r

f=ω2π=12πg5r


flag
Suggest Corrections
thumbs-up
3
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Energy in SHM
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon