JEE Advanced
Physics
Rotational Motion
2022
JEE Advanced 2022 (Paper 1)
JEE Advanced Physics Question (2022) — Solution
Question
At time t = 0 , a disk of radius 1   m starts to roll without slipping on a horizontal plane with an angular acceleration of α = 2 3   rad   s - 2 . A small stone is stuck to the disk. At t = 0 , it is at the contact point of the disk and the plane. Later, at time t = π   s , the stone detaches itself and flies off tangentially from the disk. The maximum height (in m ) reached by the stone measured from the plane is 1 2 + x 10 . The value of x is [Take g = 10   m   s - 2 .] If the numerical value has more than two decimal places, truncate/round-off the value to TWO decimal places.
Step-by-step solution
Using second equation of motion, The angle rotated by disc in t = π   s is θ = ω 0 t + 1 2 α t 2 ⇒ θ = 1 2 × 2 3 π 2 = π 3   rad Now using first equation of motion, the angular velocity of disc is ω = ω 0 + α t ⇒ ω = 0 + 2 π 3 = 2 π 3   rad   s - 1 And, velocity of disk about centre of mass is v cm = ω R = 2 π 3 × 1 = 2 π 3   m   s - 1 At the moment the stone detaches, the situation is shown in figure below. v = ω R 2 + v cm 2 + 2 ω R v cm cos 120 ° Putting the values, we get v = 2 π 3   m   s - 1 And tan θ = ω R sin 120 ° v cm + ω R cos 120 ° ⇒   tan θ = 3 ⇒   θ = π 3 rad So, the maximum height reached by the stone is H max = v 2 sin 2 θ 2 g = 2 π 3 2 × sin 2 60 ° 2 × 10 = 4 π × 3 9 × 2 × 10 × 4 = π 60   m So, height from ground will be R 1 - cos 60 ° + π 60 = 1 2 + x 10 ⇒   x = π 6 = 0 . 52
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