JEE Advanced · 2014 · 25 May 2014 · Paper 1 · Rotational Mechanics

A uniform circular disc of mass

1.5 k g

and radius

0.5 m

is initially at rest on a horizontal frictionless surface. Three forces of equal magnitude

F = 0.5 N

are applied simultaneously along the three sides of an equilateral triangle XYZ with its vertices on the perimeter of the disc (see figure). One second after applying the forces, the angular speed of the disc in

r a d s^{- 1}

Answer: 2

Solution

Each force produces a torque about the center. For a force tangent to the circle at angle, the perpendicular distance is

r sin 3 0^{\circ} = 0.5 \times 0.5 = 0.25

m.

Net torque:

τ = 3 F r sin 3 0^{\circ} = 3 \times 0.5 \times 0.5 \times 0.5 = 0.375

N·m.

Moment of inertia:

I = \frac{1}{2} m r^{2} = \frac{1}{2} \times 1.5 \times 0. 5^{2} = 0.1875

kg·m².

Angular acceleration:

α = \frac{τ}{I} = \frac{0.375}{0.1875} = 2

rad/s².

After 1 second:

ω = α t = 2 \times 1 = 2

rad/s.

JEE Advanced · 2014 · 25 May 2014 · Paper 1 · Integer answer

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JEE Advanced · 2014 · 25 May 2014 · Paper 1 · Rotational Mechanics

Solution

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