Physics, asked by TTarak111, 10 months ago

A body of mass 1 kg is revolved in a horizontal circle
by attaching to one end of 1 metre long string. If the
frequency of revolution is 20 rev/s, find about the
axis of rotation (i) moment of inertia, (ii) angular
momentum, (iii) rotational kinetic energy of the body
and (iv) centripetal force acting on the body.
Ans. (i) 1 kg mº, (ii) 40 a kg m? s?, (iii) 800 J,
(iv) 1600 N.​

Answers

Answered by iqra373
1

Answer:

Tension in the string

T = mv²/R

T = Rω²

T = 0.1 × (3)²

T = 0.1(9)

T = 0.9 N

linear velocity

V = Rω

V = 0.1(3)

V = 0.3 m/s

linear acceleration

a = v²/R

a = (0.3)²/0.1

a = 0.9 m/s²

angular acceleration

α = a/R

α = 0.9/0.1

α = 9 rad/s²

Answered by Nandhikab
0

Tension in the string

T = mv²/R

T = Rω²

T = 0.1 × (3)²

T = 0.1(9)

T = 0.9 N

linear velocity

V = Rω

V = 0.1(3)

V = 0.3 m/s

linear acceleration

a = v²/R

a = (0.3)²/0.1

a = 0.9 m/s²

angular acceleration

α = a/R

α = 0.9/0.1

α = 9 rad/s²

Similar questions