JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Rotational Mechanics

WhatsApp Email

Consider a 20 kg uniform circular disk of radius 0.2 m . It is pin supported at its centre and is at rest initially. The disk is acted upon by a constant force

F = 20 N

through a massless string wrapped around its periphery as shown in the figure. Suppose the disk makes

n

number of revolutions to attain an angular speed of

50 rad s^{- 1}

. The value of n , to the nearest integer, is ____ [Given : In one complete revolution, the disk rotates by 6.28 rad ]

Answer: 20

Solution

For a uniform disk pinned at its centre,

I = \frac{1}{2} M R^{2} = \frac{1}{2} \times 20 \times (0.2)^{2} = 0.4 kg m^{2}

. The string leaves the rim tangentially, so

τ = F R = 20 \times 0.2 = 4 N m, α = \frac{τ}{I} = \frac{4}{0.4} = 10 rad s^{- 2}

The force is constant, so the angular acceleration is constant and

ω^{2} = 2 α θ

applies from rest:

θ = \frac{( 50 ) ^{2}}{2 \times 10} = 125 rad

At 6.28 rad per revolution this is

\frac{125}{6.28} = 19.9

, so to the nearest integer

n = 20

JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Integer answer

Practise more: all JEE Main PYQs on Rotational Mechanics →

JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Rotational Mechanics

Solution

More JEE Main questions on Rotational Mechanics