A small solid sphere of mass m is released from a pt A at a height h above the bottom of a rough track as shown in fig.If the sphere rolls down the track without slipping ,its rotational kinetic energy when it comes to the bottom of track is
Asked by m.nilu
20th October 2018,
4:59 PM
Answered by Expert
Answer:
Initial energy = only potential energy = m×g×h ...............................(1)
final energy E when the ball reaches bottom = translational kinetic energy + rotational kinetic energy
E= 
.........................(2)
.........................(2)v is linear speed, ω is angular speed of rotation and r is radius of ball
By conservation of energy, if we equate (1) and (2), we get ω2 × r2 = (10/7)g×h ...........................(3)
Only rotational kinetic energy = (1/2)×I×ω2 = (1/2)×(2/5)m×r2×ω2 = (2/7)m×g×h ..............(4)
Eqn.(3) is used to substitute ω2 × r2 in (4)
Answered by Expert
21st October 2018,
10:11 PM
Rate this answer
- 1
- 2
- 3
- 4
- 5
- 6
- 7
- 8
- 9
- 10
You have rated this answer /10
Your answer has been posted successfully!
RELATED STUDY RESOURCES :
Browse free questions and answers by Chapters
- 1 Thermodynamics
- 2 Gravitation
- 3 Electromagnetic Waves
- 4 Communication Systems
- 5 Laws of Motion
- 6 Current Electricity
- 7 Work, Energy and Power
- 8 Kinematics
- 9 Physics and Measurement
- 10 Rotational Motion
- 11 Properties of Solids and Liquids
- 12 Kinetic Theory of Gases
- 13 Oscillations and Waves
- 14 Electrostatics
- 15 Magnetic Effects of Current and Magnetism
- 16 Electromagnetic Induction and Alternating Currents
- 17 Optics
- 18 Dual Nature of Matter and Radiation
- 19 Atoms and Nuclei
- 20 Electronic Devices
Trending Tags
Latest Questions
JEE Main Mathematics
Asked by amaladelsingh
28th October 2020,
10:24 PM
CBSE X Biology
Asked by reenacharan2008
28th October 2020,
9:11 PM
