Banner 0
Banner 1
Banner 2
Banner 3
Banner 4
Banner 5
Banner 6

Q.

A semicircular track of radius R=62.5 cm is cut in a block. Mass of block, having track, is M=1 kg and rests over a smooth horizontal floor. A cylinder of radius r=10 cm and mass m=0.5 kg is hanging by thread such that axes of cylinder and track are in same level and surface of cylinder is in contact with the track as shown in figure. When the thread is burnt, cylinder starts to move down the track. Sufficient friction exists between surface of cylinder and track, so that cylinder does not slip. Find force  (in Newton) applied by block on the floor at that moment. g=10 m/s2

A semicircular track of radius R = 62.5 cm is cut in a block .Mass of block  ,having track,is M = 1 kg and rests over a smooth horizontal floor.A  cylinder of

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away
An Intiative by Sri Chaitanya

a

21.57 N

b

15.57 N

c

10.57 N

d

13.57 N 

answer is D.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

From conservational of linear momentum

Question Image

mv1=Mv2

 Velocity of cylinder axis relative to block

vr=v1+v2

Applying conservation of mechanical energy,

mg(R-r)=12mv12+122+12Mv22

Here, I=12mr2   and   ω=vrr

Solving the above equations with given data. we get

v1=2.0 m/s v2=1.0 m/s Further, N-mg=mv2R-r N=mg+mvr2R-r=(0.5)(10)+(0.5)(3.0)20.525 =13.57 N

Watch 3-min video & get full concept clarity

Best Courses for You

JEE

JEE

NEET

NEET

Foundation JEE

Foundation JEE

Foundation NEET

Foundation NEET

CBSE

CBSE

score_test_img

Get Expert Academic Guidance – Connect with a Counselor Today!

whats app icon