Question

A uniform rod of mass 300 g and length 50 cm rotates at a uniform angular speed of 2 rad/s about an axis perpendicular to the rod through an end. Calculate (a) the angular momentum of the rod about the axis of rotation, (b) the speed of the centre of the rod and (c) its kinetic energy.

Answer 1

• The angular momentum of the rod about its axis of rotation is 0.05 kg m²/sec.
• The speed of the center of rod is 0.5 m/sec
• Total kinetic energy of the rod is 0.05 Joule.

Given

Mass of rod = 300 g

Length of rod = 50 cm

Angular speed = 2 rad/sec

a) We know that angular momentum L = Iω where I is the moment of inertia and ω is the angular speed.

By putting the values we get-

L = Iω = ml²/3 × ω

L = 0.3 × (0.5)² / 3 × 2 = 0.05 kg m²/sec

b) To find the speed of the center of the rod

V = rω

V = 0.25 × 2

V = 0.5 m/sec

c) To find the total kinetic energy of the rod-

Here rod is in pure rotational motion so kinetic energy is equals to rotational kinetic energy.

KE = 1/2 Iω²

KE = 1/2 × 0.025 × 2× 2

KE = 0.05 J

Regards

Answer 2

Answer:

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