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

If a solid sphere of mass 1 kg and radius 0.1 m rolls without slipping at a uniform velocity of 1 m/s along a straight line on a horizontal floor, then its kinetic energy is

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

The correct option is C 710 J
When a body is in state of pure rolling te condition which hold is (v=rω) and it has translational K.E as well as rotational K.E
K.Etotal=K.ETrans+K.ERot=12mv2CM+12ICMω2 ....(i)
Where v=vCM, ω=vr and ICM=25mr2 for solid sphere.
From Eq. (i),
K.Etotal=12mv2+[12(25mr2)×v2r2]
K.Etotal=12mv2+15mv2=710mv2
K.Etotal=710×1×12=710 J

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