Question

The two particles of mass m and 2m, respectively are connected by a light rod of negligible ma slide with negligible friction on a circular path of radius r on the inside of a fixed vertical circular the system is released from rest at $\theta$ = 0° and $\theta$ is taken from positive x-axis in clockwise directi

• A. The speed of the particles when the rod passes the horizontal position is $\sqrt{\frac{2gr}{3}(\frac{3}{\sqrt{2}}-1)}$
• B. The maximum speed of the particles is $\sqrt{\frac{2}{3}gr(\sqrt{5}-1)}$
• C. The maximum speed of the particles is at $\theta$ = $\tan^{-1}(\frac{1}{2})$
• D. The maximum value of $\theta$ is $2\tan^{-1}(2)$.

Answer 1

There is a fundamental inconsistency in the problem statement regarding the definition of θ and the initial position θ=0 in relation to the diagram. None of the options match.

Answer 2

The problem describes two particles of masses m and 2m connected by a light rod, sliding without friction on a circular path of radius r inside a fixed vertical circular track. The system is released from rest at θ = 0°, where θ is taken from the positive x-axis in a clockwise direction. The diagram shows the rod connecting two points on the circle that are 90 degrees apart. This implies the length of the rod is L = r√2.

1. Initial Configuration and Potential Energy:

Let the center of the circular track be the origin (0,0). The problem states θ = 0 is the initial position, and θ is measured clockwise from the positive x-axis. The diagram shows θ as the angle of the rod with the horizontal. For the rod to be horizontal and connect two points on the circle separated by 90 degrees, the particles must be at:

• Mass m: (-r/√2, r/√2) (angle 3π/4 or 135° counter-clockwise from positive x-axis).
• Mass 2m: (r/√2, r/√2) (angle π/4 or 45° counter-clockwise from positive x-axis).

Initial y-coordinates: y_{m,initial} = r/√2 and y_{2m,initial} = r/√2. Initial Potential Energy (taking the center of the circle as reference y=0): $PE_{initial} = m g y_{m,initial} + 2m g y_{2m,initial} = m g (r/√2) + 2m g (r/√2) = 3m g r/√2$ Initial Kinetic Energy: $KE_{initial} = 0$ (released from rest). Total Initial Energy: $E_{initial} = 3m g r/√2$.

2. Configuration at Angle θ and Energy:

Let the rod rotate clockwise by an angle θ from its initial horizontal position. The new angular positions (counter-clockwise from positive x-axis) are:

• Mass m: α_m = 3π/4 - θ
• Mass 2m: α_{2m} = π/4 - θ

The new y-coordinates are: $y_m = r \sin(3π/4 - θ) = r (\sin(3π/4)\cosθ - \cos(3π/4)\sinθ) = r ( (1/√2)\cosθ - (-1/√2)\sinθ ) = (r/√2)(\cosθ + \sinθ)$ $y_{2m} = r \sin(π/4 - θ) = r (\sin(π/4)\cosθ - \cos(π/4)\sinθ) = r ( (1/√2)\cosθ - (1/√2)\sinθ ) = (r/√2)(\cosθ - \sinθ)$

Potential Energy at angle θ: $PE(\theta) = m g y_m + 2m g y_{2m} = m g (r/√2)(\cosθ + \sinθ) + 2m g (r/√2)(\cosθ - \sinθ)$ $PE(\theta) = (mgr/√2) [\cosθ + \sinθ + 2\cosθ - 2\sinθ] = (mgr/√2) [3\cosθ - \sinθ]$

Since the rod is rigid and the particles move on a circle of radius r, their speeds are the same, v = rω, where ω is the angular velocity of the rod. Kinetic Energy at angle θ: $KE(\theta) = (1/2) m v^2 + (1/2) (2m) v^2 = (1/2) (3m) v^2$

3. Conservation of Mechanical Energy: $KE_{initial} + PE_{initial} = KE(\theta) + PE(\theta)$ $0 + 3m g r/√2 = (3/2) m v^2 + (mgr/√2) [3\cosθ - \sinθ]$ Divide by m: $3 g r/√2 = (3/2) v^2 + (gr/√2) [3\cosθ - \sinθ]$ $(3/2) v^2 = (gr/√2) [3 - (3\cosθ - \sinθ)]$ $(3/2) v^2 = (gr/√2) [3 - 3\cosθ + \sinθ]$ $v^2 = (2/3) (gr/√2) [3 - 3\cosθ + \sinθ]$ $v^2 = (gr√2/3) [3 - 3\cosθ + \sinθ]$

4. Evaluate the Options:

(A) The speed of the particles when the rod passes the horizontal position: The rod is horizontal at θ = 0 (initial position, $v=0$) and at θ = π (180° clockwise rotation). At θ = π: $\cos(π) = -1$, $\sin(π) = 0$. $v^2 = (gr√2/3) [3 - 3(-1) + 0] = (gr√2/3) [3 + 3] = (gr√2/3) \times 6 = 2gr√2$ $v = \sqrt{2gr√2}$ Option (A) is $\sqrt{\frac{2gr}{3}(\frac{3}{\sqrt{2}}-1)} = \sqrt{\frac{2gr}{3}(\frac{3\sqrt{2}}{2}-1)} = \sqrt{gr(\sqrt{2}-2/3)}$. This does not match $\sqrt{2gr√2}$. So, option (A) is incorrect.

(B) The maximum speed of the particles: Speed v is maximum when v^2 is maximum. Let $f(\theta) = 3 - 3\cosθ + \sinθ$. To find the maximum of $f(\theta)$, set its derivative to zero: $f'(\theta) = 3\sinθ + \cosθ = 0$ $3\sinθ = -\cosθ \implies \tanθ = -1/3$

Since the system starts from θ=0 and moves downwards, θ will be positive. $\tanθ = -1/3$ means θ is in the second or fourth quadrant. For the system to gain speed, its potential energy must decrease, meaning it moves to a lower position. This implies θ is in the fourth quadrant (e.g., θ between 0 and 2π).

If $\tanθ = -1/3$, then we can construct a right triangle with opposite side 1 and adjacent side 3. The hypotenuse is $\sqrt{1^2 + 3^2} = \sqrt{10}$. For θ in the fourth quadrant: $\sinθ = -1/\sqrt{10}$ and $\cosθ = 3/\sqrt{10}$. Substitute these values into the expression for $v^2$: $v^2_{max} = (gr√2/3) [3 - 3(3/\sqrt{10}) + (-1/\sqrt{10})]$ $v^2_{max} = (gr√2/3) [3 - 9/\sqrt{10} - 1/\sqrt{10}]$ $v^2_{max} = (gr√2/3) [3 - 10/\sqrt{10}] = (gr√2/3) [3 - \sqrt{10}]$ $v^2_{max} = (gr√2/3) [3 - \sqrt{10}]$.

Option (B) is $\sqrt{\frac{2}{3}gr(\sqrt{5}-1)}$. This is $\sqrt{\frac{2gr}{3}(\sqrt{5}-1)}$. Our result is $v_{max} = \sqrt{\frac{gr\sqrt{2}}{3}(3-\sqrt{10})}$.

There is a fundamental inconsistency in the problem statement regarding the definition of θ and the initial position θ=0 in relation to the diagram.