A body of mass 1kg tied to one end of string is revolved in a horizontal circle of radius 0.1 m with a speed of 3 revolution /sec , assuming the effect of gravity is negligible , then linear velocity , acceleration and tension in the string will be
1. 1.88m/s, 35.5m/^2 , 35.5N
2.2.88m/s, 45.5m/s^2, 45.5N
3.3.88m/s, 55.5m/s^2, 55.5N
4.None of these
Answers
Answered by
33
option 1.......linear velocity will be 0.6*pi=1.88m/s....as angular velocity 6*pi rad/sec...multiply with radius to get that...linear acceleration is 35.5 m/s^2 as angular acceleration is ( linear velocity^2/radius)....and multiply it with radius again to get linear acceleration... as 3.6*(pi)^2=35.5......and tension is (m(v^2)/r) ie centrifugal force ...=35.5N..
Answered by
11
Answer:
option 1
step by step explanation
- omega=theta/time
- omega=2π*3/1
- omega=6π
- velocity=omega*radius
- velocity=6π*0.1
- velocity= 1.88m/s
- acceleration=omega^2*radius
- acceleration=36π^2*0.1
- acceleration=35.5m/s^2
- Tension=mass*velocity^2/radius
- tension=1*1.88*1.88/0.1
- Tension=35.5N
HOPE THIS EXPLANATION WILL BE HELPFUL TO YOU
DON'T FORGET TO MAKE THIS ANSWER AS BRAINLIEST
Similar questions