WPE O12

Solution 12

Solution to Question 12

θ Force on Tube

1. Velocity of the Ball

Let $R$ be the radius. From conservation of energy, the potential energy lost by the ball as it falls through an angle $\theta$ (measured from the vertical) is converted into kinetic energy.

$$ mgR(1 – \cos\theta) = \frac{1}{2}mv^2 \implies v^2 = 2gR(1 – \cos\theta) $$

2. Radial Force Analysis

Consider the forces on the ball in the radial direction (towards the center). Let $N_{in}$ be the normal force exerted by the outer wall on the ball (directed inwards). The equation of motion is:

$$ N_{in} + mg \cos\theta = \frac{mv^2}{R} $$ $$ N_{in} = \frac{m}{R}[2gR(1 – \cos\theta)] – mg \cos\theta $$ $$ N_{in} = mg(2 – 2\cos\theta – \cos\theta) = mg(2 – 3\cos\theta) $$

Force exerted by the ball on the tube: By Newton’s 3rd Law, the ball exerts a force equal and opposite to $N_{in}$ on the tube. This force is directed radially outwards with magnitude $F_{out} = mg(2 – 3\cos\theta)$. (Note: If this value is negative, the force is directed inwards, which physically means the ball presses against the inner wall).

3. Vertical Force on the Floor

The total normal reaction $N_f$ from the floor supports the weight of the tube $Mg$ and the vertical component of the forces exerted by the two balls.

The vertical component of the outward radial force $F_{out}$ is $F_y = F_{out} \cos\theta$ (directed upwards). Let’s verify direction. The radial vector points down/out. The outward force points up/out from the center. Its vertical component is $F_{out} \cos\theta$ (upwards).

Force on tube (Vertical) = $F_{out} \cos\theta = mg(2 – 3\cos\theta)\cos\theta$ (Upwards).

The normal reaction from the ground $N_f$ balances the downward weight of the tube and the vertical forces from the balls.

$$ N_f + 2(F_{out} \cos\theta) – Mg = 0 $$ $$ N_f = Mg – 2mg(2\cos\theta – 3\cos^2\theta) $$ $$ N_f = Mg + 2mg(3\cos^2\theta – 2\cos\theta) $$

4. Analyzing the Variation

Let $x = \cos\theta$. As the ball moves down, $x$ goes from $1 \to 0$. We analyze the function $y(x) = 3x^2 – 2x$.

  • At start ($\theta = 0^\circ, x=1$): $y = 3(1) – 2 = 1$. ($N_f = Mg + 2mg$).
  • The minimum occurs at $\frac{dy}{dx} = 6x – 2 = 0 \implies x = 1/3$.
  • At minimum ($x=1/3$): $y = 3(1/9) – 2(1/3) = -1/3$. ($N_f = Mg – \frac{2}{3}mg$).
  • At end ($\theta = 90^\circ, x=0$): $y = 0$. ($N_f = Mg$).

The value of $N_f$ starts at a maximum, decreases to a minimum, and then increases. This matches option (c).

Note on Option (d): If the tube is light enough such that $Mg < \frac{2}{3}mg$, the normal force $N_f$ would become negative, meaning the tube would lift off the ground.

Correct Option: (c) and (d)

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