Question

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Answer 1

hope this will help you.....

Answer 2

Answer:

Option "4" Is Al Right!!

Explanation:

Given:

A solid cylinder of mass 2 kg and radius 4 cm rotating about its axis at the rate of 3 rpm. The torque required to stop after 2π revolutions is

(1) 2 × 10–6 N m

(2) 2 × 10–3 N m

(3) 12 × 10–4 N m

(4) 2 × 106 N m

Solution:

Mass of solid cylinder , m = 2kg

Radius of cylinder , R = 4cm = 0.04m

A solid cylinder is rotating about its axis,

So, momentof inertia, I = 1/2 mr²

= 1/2 × 2 × (0.04)² = 16 × 10^-4 Kgm²

= 1.6 × 10^-3 kgm²

Angular frequency,  = 3rpm = 3 × 2π/60 rad/s = π/10 rad/s

Angular displacement = 2π revolution

= 2π × 2π = 4π² [ as 1 revolution = 2π ]

From conservation of energytheorem,

Change in rotational kinetic energy = torque × angular displacement

Or, 1/2 I = torque × 4π²

Or, 1/2 × 1.6 × 10^-3 × 1/100 = torque × 4

Torque = 2 × 10^-6 Nm

Hence = 2×10^-6Nm

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The Answer is option is "4" = 2*10^-6  Nm