Hey there! We receieved your request
Stay Tuned as we are going to contact you within 1 Hour
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-5470-145
+91 7353221155
Use Coupon: CART20 and get 20% off on all online Study Material
Complete Your Registration (Step 2 of 2 )
Sit and relax as our customer representative will contact you within 1 business day
OTP to be sent to Change
A BULLET IS FIRED HORIZONTALLY WITH A SPEED OF V TO GRAGE OUT A SOLID CYLINDER PLACED ON A ROUGH SURFACE. AFTER COLLISION , THE BULLET MOVED TANGENTIALLY AT THE TOP AND THE CYLINDER ROLLS WITHOUT SLIPPING. IF THE MASS OF CYLINDER IS M , RADIUS R THE ANGULAR VELOCITY OF THE CYLINDER IS
Applying conservation of angular momentum about the point of contact between cylinder & rough surface taking both bullet & cylinder as system,
mV*2R=(3/2*MR2)*ω m=mass of bullet,Moment of inertia I(about the axis passing through contact point & || to
axis of cylinder)=3/2*MR2 (by applying || axis theorem)
or, ω=4/3*(m/M)*(V/R)
Note:Judicious choice of point of application eliminate frictional torque(external to the system) here.
IN THE BULLET PROBLEM WHAT ABOUT THE VELOCITY OF BULLET AFTER COLLISION AND U DID NOT TAKE BULLET ANGULAR MOMENTUM AFTER COLLISION WHY
PL POST THE ANSWER
Register Yourself for a FREE Demo Class by Top IITians & Medical Experts Today !