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Question: A solid cylinder is kept on one edge of a plank of same mass and length 25 m placed on a smooth surf...

A solid cylinder is kept on one edge of a plank of same mass and length 25 m placed on a smooth surface as shown in the figure. The coefficient of friction between the cylinder and the plank is 0. 5. The plank is given a velocity of 20 m s−1^{-1} towards right. Find the time (in s) after which plank and cylinder will separate. [g = 10 m s−2^{-2}]

Answer

Separation occurs after approximately 6.45 seconds.

Explanation

Solution

Solution

We follow the method given in the similar problem. When the plank is suddenly set in motion with speed 20 m/s and the cylinder (which is in contact with one edge) has not received any kick, kinetic friction acts between them. Taking

   fₖ = μmg = 0.5·m·10 = 5 N

the translational accelerations (assuming both bodies have the same mass m) are

   a_cylinder = +5 m/s²    (to the right)
   a_plank = –5 m/s²     (to the right; friction on the plank opposes its motion)

In the reference frame of the plank (taking “separation” as the instant when the cylinder has rolled across the full plank length L), we may write the kinematic equation for the distance s along the plank that must be covered by the cylinder. In the similar question the equation was taken as

   s = u t + (1⁄2)(–5)t²

where u = 20 m/s (the initial relative speed between the cylinder and plank) and s is the length of the plank. In our problem the plank length is L = 25 m so that

   25 = 20 t – 2.5 t²

This can be rearranged as

   2.5t² – 20t + 25 = 0  ⟹  t² – 8t + 10 = 0

Solve using the quadratic formula:

   t = [8 ± √(64 – 40)]⁄2 = [8 ± √24]⁄2 = 4 ± √6

There are two positive roots. As in the similar problem the later time corresponds to the instant when the cylinder finally leaves the plank. Therefore, the separation time is

   t = 4 + √6  ≈ 4 + 2.45 = 6.45 s

Explanation (Minimal)

  1. Friction force fₖ = μmg = 5 N.
  2. Using Newton’s 2nd law the effective deceleration (taken from the similar problem’s approach) is 5 m/s² acting against the initial relative speed of 20 m/s.
  3. Taking s = 25 m, we set    25 = 20t – (1/2)(5)t² = 20t – 2.5t²    ⟹ t² – 8t + 10 = 0
  4. Solving gives t = 4 ± √6; the physically relevant (later) time is t = 4 + √6 ≈ 6.45 s.

Answer: Separation occurs after approximately 6.45 seconds. (Option: If multiple‐choice, choose the option with 6.45 s.)